Explanation:
Okay, let's try to figure out how many students take physics. There are 360 people in total. Some take calculus, physics, chemistry, or none. Let me jot down the given information first.

First, 15 students take all three subjects: calculus, physics, and chemistry. Then, 15 don't take any of them. So, the number of students taking at least one of the subjects is 360 - 15 = 345.

Next, 180 take calculus. So, the total number of students in calculus is 180. Also, twice as many students take chemistry as take physics. Let me note that as C = 2P, where C is the number taking chemistry and P is the number taking physics.

Then, there are some overlaps: 75 take both calculus and chemistry, 75 take both physics and chemistry, and only 30 take both physics and calculus. Wait, the problem says "Only 30 take both physics and calculus." So, the intersection of physics and calculus is 30. But earlier, it said 15 take all three. So, does that mean that the number taking both physics and calculus is 30, which includes those who also take chemistry? So, the 30 includes the 15 who take all three? Let me confirm.

Yes, when they say "Only 30 take both physics and calculus," that probably means that the intersection of physics and calculus is 30, which includes those who might also be taking chemistry. But since 15 take all three, that would mean that the number taking both physics and calculus but not chemistry is 30 - 15 = 15? Wait, but maybe the problem is using "only" to mean that there are no other overlaps? Wait, no, the wording is a bit confusing here. Let me check again.

The problem says: "Only 30 take both physics and calculus." Hmm. The word "only" here might be indicating that the number of students taking both physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement that 15 take all three. So, maybe the problem is using "only" to mean that the number taking both physics and calculus, excluding those taking all three, is 30? Wait, that might not be standard. Let me think.

In standard set theory, when they say "the number taking both physics and calculus is 30," that includes those taking all three. But if they say "only 30 take both physics and calculus," that might mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in a different way. Wait, maybe the problem is saying that the number taking both physics and calculus, but not chemistry, is 30? That would make sense. Let me check the original problem again.

Original problem: "Only 30 take both physics and calculus." So, maybe "only" here is meant to indicate that there are no students taking all three? But that conflicts with the first statement. Wait, the first statement says 15 take all three. So, perhaps the problem is using "only" to mean that the number taking both physics and calculus is 30, and that there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in the sense that the number taking both physics and calculus, excluding those taking all three, is 30? That is, the intersection of physics and calculus is 30 + 15 = 45? Wait, but the problem says "Only 30 take both physics and calculus," so maybe the total number taking both physics and calculus is 30, which includes those taking all three? But then how does that reconcile with the 15 taking all three?

This is confusing. Let me try to parse the problem again:

"15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?"

So, the key is that "Only 30 take both physics and calculus." So, the number of students taking both physics and calculus is 30. However, there are 15 students taking all three. Therefore, the number of students taking both physics and calculus but not chemistry is 30 - 15 = 15. Is that correct? Because the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the overlap between physics and calculus is 30, which includes the 15 who take all three. Therefore, the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Similarly, the number taking both calculus and chemistry is 75. Since 15 take all three, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

So, with that in mind, let me try to use the principle of inclusion-exclusion for three sets. Let me denote:

Let C be the number taking calculus, P the number taking physics, Ch the number taking chemistry.

We know that:

Total students taking at least one subject: 345 (since 360 - 15 = 345)

We have:

C = 180

Ch = 2P (since twice as many take chemistry as take physics)

Also, the number taking both calculus and chemistry is 75, which includes those taking all three. Similarly, the number taking both physics and chemistry is 75, and the number taking both physics and calculus is 30, which includes those taking all three.

So, using inclusion-exclusion formula for three sets:

Total = C + P + Ch - (number taking two subjects) + (number taking all three)

Wait, the formula is:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C ∩ P| - |C ∩ Ch| - |P ∩ Ch| + |C ∩ P ∩ Ch|

So, substituting the known values:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute that step by step.

First, let me note that |C ∩ P| is 30, |C ∩ Ch| is 75, |P ∩ Ch| is 75, and |C ∩ P ∩ Ch| is 15.

So:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute the constants:

180 - 30 - 75 - 75 + 15 = 180 - 30 is 150; 150 - 75 is 75; 75 - 75 is 0; 0 + 15 is 15. So:

345 = 15 + P + Ch

Therefore, P + Ch = 345 - 15 = 330.

But we also know that Ch = 2P. Therefore, substituting:

P + 2P = 330 => 3P = 330 => P = 110.

Therefore, the number of students taking physics is 110. Wait, but let me check if this is consistent with other numbers.

Wait, let me verify. If P = 110, then Ch = 220. Let me check if the inclusion-exclusion formula holds.

So, |C ∪ P ∪ Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = let's compute:

180 + 110 + 220 = 510

Then subtract 30 + 75 + 75 = 180, so 510 - 180 = 330

Then add back 15: 330 + 15 = 345. Which matches the total. So that seems correct.

But let me check if the numbers for the overlaps make sense. For example, the number taking both calculus and chemistry is 75. Which includes those taking all three. So, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60. The number taking both physics and calculus is 30, which includes those taking all three, so the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Now, let's check if the total number of students taking each subject is consistent.

For calculus: The total number taking calculus is 180. This includes:

- Those taking only calculus.

- Those taking calculus and physics but not chemistry: 15.

- Those taking calculus and chemistry but not physics: 60.

- Those taking all three: 15.

So, total taking calculus is 15 (only calculus) + 15 (calculus and physics) + 60 (calculus and chemistry) + 15 (all three) = 15 + 15 + 60 + 15 = 105? Wait, but that's supposed to be 180. Wait, that can't be. There's a problem here. Wait, so according to this, the total number taking calculus would be 105, but we know it's 180. So, clearly, something is wrong here.

Wait, so my mistake must be in how I calculated the numbers. Let me think again.

Wait, the formula for the total number taking calculus is:

Number taking only calculus + number taking calculus and physics only + number taking calculus and chemistry only + number taking all three.

But according to the problem, the number taking both calculus and physics is 30, which includes those taking all three. So, the number taking calculus and physics only is 30 - 15 = 15. Similarly, the number taking calculus and chemistry is 75, which includes those taking all three, so the number taking calculus and chemistry only is 75 - 15 = 60. Then, the number taking all three is 15. So, the total number taking calculus is:

Only calculus + 15 (calculus and physics only) + 60 (calculus and chemistry only) + 15 (all three) = ?

But we know that total taking calculus is 180. Therefore:

Only calculus + 15 + 60 + 15 = Only calculus + 90 = 180 => Only calculus = 90.

So, the number taking only calculus is 90.

Similarly, let's check for physics. The total number taking physics is P = 110. Which includes:

Only physics + physics and calculus only + physics and chemistry only + all three.

Which is:

Only physics + 15 (physics and calculus only) + 60 (physics and chemistry only) + 15 (all three) = Only physics + 15 + 60 + 15 = Only physics + 90 = 110 => Only physics = 20.

Similarly, for chemistry, total number taking chemistry is Ch = 220. Which includes:

Only chemistry + calculus and chemistry only + physics and chemistry only + all three.

Which is:

Only chemistry + 60 (calculus and chemistry only) + 60 (physics and chemistry only) + 15 (all three) = Only chemistry + 60 + 60 + 15 = Only chemistry + 135 = 220 => Only chemistry = 85.

Now, let's check the total number of students taking at least one subject:

Only calculus (90) + Only physics (20) + Only chemistry (85) + calculus and physics only (15) + calculus and chemistry only (60) + physics and chemistry only (60) + all three (15) =

Let me add them up:

90 + 20 = 110

110 + 85 = 195

195 + 15 = 210

210 + 60 = 270

270 + 60 = 330

330 + 15 = 345.

Which matches the total of 345. So that seems okay.

But then, the problem is that when I first thought about the total number taking calculus, I thought that the numbers didn't add up, but actually, there is a "only calculus" component which is 90. So, that's okay.

So, according to this, the number of students taking physics is 110. But let me check if there are any other constraints or if I made a wrong assumption.

Wait, the problem says "Twice as many students take chemistry as take physics." So, Ch = 2P. Since we found P = 110, then Ch = 220, which matches our calculation.

Also, the number taking both calculus and chemistry is 75, which is given. The number taking both physics and chemistry is 75, which is given. The number taking both physics and calculus is 30, which is given. The number taking all three is 15.

So, all the numbers check out. Therefore, the answer should be 110.

But let me just think again if there's another way to approach this problem, maybe using a Venn diagram or another method, to confirm.

Alternatively, let's use variables for the different regions in the Venn diagram. Let me define:

Let a = only calculus

b = only physics

c = only chemistry

d = calculus and physics only

e = calculus and chemistry only

f = physics and chemistry only

g = all three

We know that:

Total students: 360

Students taking none: 15

Therefore, students taking at least one: 345 = a + b + c + d + e + f + g

Given:

g = 15

Number taking calculus: a + d + e + g = 180

Number taking physics: b + d + f + g = P

Number taking chemistry: c + e + f + g = Ch = 2P

Number taking both calculus and chemistry: e + g = 75 => e = 75 - g = 75 - 15 = 60

Number taking both physics and chemistry: f + g = 75 => f = 75 - 15 = 60

Number taking both physics and calculus: d + g = 30 => d = 30 - 15 = 15

So, we have:

d = 15, e = 60, f = 60, g = 15

Now, let's plug into the equation for calculus:

a + d + e + g = a + 15 + 60 + 15 = a + 90 = 180 => a = 90

Similarly, for chemistry:

c + e + f + g = c + 60 + 60 + 15 = c + 135 = 2P => c = 2P - 135

For physics:

b + d + f + g = b + 15 + 60 + 15 = b + 90 = P => b = P - 90

Now, total students taking at least one:

a + b + c + d + e + f + g = 90 + (P - 90) + (2P - 135) + 15 + 60 + 60 + 15

Let me compute this:

First, expand the terms:

90 + P - 90 + 2P - 135 + 15 + 60 + 60 + 15

Simplify step by step:

Start with 90 - 90 = 0

Then, 0 + P + 2P = 3P

Then, 3P - 135 + 15 + 60 + 60 + 15

Compute constants:

-135 + 15 = -120

-120 + 60 = -60

-60 + 60 = 0

0 + 15 = 15

So overall, total is 3P + 15

But we know that total is 345, so:

3P + 15 = 345 => 3P = 330 => P = 110

So, same result. Therefore, the number of students taking physics is 110.

Therefore, the answer is 110.

But let me check once more if there's any inconsistency. For example, the number taking only chemistry is c = 2P - 135 = 2*110 - 135 = 220 - 135 = 85. Which is positive, so that's okay. The number taking only physics is b = P - 90 = 110 - 90 = 20. Also positive.

So, all numbers are positive, which is good.

Therefore, I think the answer is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine how many students take **physics**, we are given a set of constraints involving three subjects: **calculus**, **physics**, and **chemistry**, and the total number of students in the school. Let's break this down step by step using the **principle of inclusion-exclusion** and logical deductions.

---

### **Step 1: Total Students and Those Taking No Subjects**

- Total students in the school: **360**
- Students who take **none** of the three subjects: **15**
- Therefore, students taking **at least one** of the subjects:
  $$
  360 - 15 = 345
  $$

---

### **Step 2: Define Variables and Known Values**

Let:
- $ C $ = number of students taking **calculus** = **180**
- $ P $ = number of students taking **physics** = **unknown**
- $ Ch $ = number of students taking **chemistry** = **unknown**
- $ |C \cap P| $ = number taking **both calculus and physics** = **30**
- $ |C \cap Ch| $ = number taking **both calculus and chemistry** = **75**
- $ |P \cap Ch| $ = number taking **both physics and chemistry** = **75**
- $ |C \cap P \cap Ch| $ = number taking **all three** = **15**

Also, it's given that:
$$
Ch = 2P
$$

---

### **Step 3: Apply Inclusion-Exclusion Principle**

The formula for the number of students taking **at least one** of the three subjects is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

$$
345 = 180 + P + Ch - 30 - 75 - 75 + 15
$$

Simplify the constants:

$$
345 = 180 + P + Ch - 180 + 15 = P + Ch + 15
$$

So:

$$
P + Ch = 330
$$

---

### **Step 4: Use the Given Relationship Between Chemistry and Physics**

We are told that:

$$
Ch = 2P
$$

Substitute into the equation:

$$
P + 2P = 330 \Rightarrow 3P = 330 \Rightarrow P = 110
$$

---

### **Step 5: Verify the Solution**

- $ P = 110 $
- $ Ch = 2P = 220 $
- Check the inclusion-exclusion formula again:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This matches the total number of students taking at least one subject, confirming the solution is consistent.

---

### **Final Answer**

$$
\boxed{110}
$$

[Assistant]
Explanation:
Okay, let's try to figure out how many students take physics. There are 360 people in total. Some take calculus, physics, chemistry, or none. Let me jot down the given information first.

First, 15 students take all three subjects: calculus, physics, and chemistry. Then, 15 don't take any of them. So, the number of students taking at least one of the subjects is 360 - 15 = 345.

Next, 180 take calculus. So, the total number of students in calculus is 180. Also, twice as many students take chemistry as take physics. Let me note that as C = 2P, where C is the number taking chemistry and P is the number taking physics.

Then, there are some overlaps: 75 take both calculus and chemistry, 75 take both physics and chemistry, and only 30 take both physics and calculus. Wait, the problem says "Only 30 take both physics and calculus." So, the intersection of physics and calculus is 30. But earlier, it said 15 take all three. So, does that mean that the number taking both physics and calculus is 30, which includes those who also take chemistry? So, the 30 includes the 15 who take all three? Let me confirm.

Yes, when they say "Only 30 take both physics and calculus," that probably means that the intersection of physics and calculus is 30, which includes those who might also be taking chemistry. But since 15 take all three, that would mean that the number taking both physics and calculus but not chemistry is 30 - 15 = 15? Wait, but maybe the problem is using "only" to mean that there are no other overlaps? Wait, no, the wording is a bit confusing here. Let me check again.

The problem says: "Only 30 take both physics and calculus." Hmm. The word "only" here might be indicating that the number of students taking both physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement that 15 take all three. So, maybe the problem is using "only" to mean that the number taking both physics and calculus, excluding those taking all three, is 30? Wait, that might not be standard. Let me think.

In standard set theory, when they say "the number taking both physics and calculus is 30," that includes those taking all three. But if they say "only 30 take both physics and calculus," that might mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in a different way. Wait, maybe the problem is saying that the number taking both physics and calculus, but not chemistry, is 30? That would make sense. Let me check the original problem again.

Original problem: "Only 30 take both physics and calculus." So, maybe "only" here is meant to indicate that there are no students taking all three? But that conflicts with the first statement. Wait, the first statement says 15 take all three. So, perhaps the problem is using "only" to mean that the number taking both physics and calculus is 30, and that there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in the sense that the number taking both physics and calculus, excluding those taking all three, is 30? That is, the intersection of physics and calculus is 30 + 15 = 45? Wait, but the problem says "Only 30 take both physics and calculus," so maybe the total number taking both physics and calculus is 30, which includes those taking all three? But then how does that reconcile with the 15 taking all three?

This is confusing. Let me try to parse the problem again:

"15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?"

So, the key is that "Only 30 take both physics and calculus." So, the number of students taking both physics and calculus is 30. However, there are 15 students taking all three. Therefore, the number of students taking both physics and calculus but not chemistry is 30 - 15 = 15. Is that correct? Because the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the overlap between physics and calculus is 30, which includes the 15 who take all three. Therefore, the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Similarly, the number taking both calculus and chemistry is 75. Since 15 take all three, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

So, with that in mind, let me try to use the principle of inclusion-exclusion for three sets. Let me denote:

Let C be the number taking calculus, P the number taking physics, Ch the number taking chemistry.

We know that:

Total students taking at least one subject: 345 (since 360 - 15 = 345)

We have:

C = 180

Ch = 2P (since twice as many take chemistry as take physics)

Also, the number taking both calculus and chemistry is 75, which includes those taking all three. Similarly, the number taking both physics and chemistry is 75, and the number taking both physics and calculus is 30, which includes those taking all three.

So, using inclusion-exclusion formula for three sets:

Total = C + P + Ch - (number taking two subjects) + (number taking all three)

Wait, the formula is:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C ∩ P| - |C ∩ Ch| - |P ∩ Ch| + |C ∩ P ∩ Ch|

So, substituting the known values:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute that step by step.

First, let me note that |C ∩ P| is 30, |C ∩ Ch| is 75, |P ∩ Ch| is 75, and |C ∩ P ∩ Ch| is 15.

So:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute the constants:

180 - 30 - 75 - 75 + 15 = 180 - 30 is 150; 150 - 75 is 75; 75 - 75 is 0; 0 + 15 is 15. So:

345 = 15 + P + Ch

Therefore, P + Ch = 345 - 15 = 330.

But we also know that Ch = 2P. Therefore, substituting:

P + 2P = 330 => 3P = 330 => P = 110.

Therefore, the number of students taking physics is 110. Wait, but let me check if this is consistent with other numbers.

Wait, let me verify. If P = 110, then Ch = 220. Let me check if the inclusion-exclusion formula holds.

So, |C ∪ P ∪ Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = let's compute:

180 + 110 + 220 = 510

Then subtract 30 + 75 + 75 = 180, so 510 - 180 = 330

Then add back 15: 330 + 15 = 345. Which matches the total. So that seems correct.

But let me check if the numbers for the overlaps make sense. For example, the number taking both calculus and chemistry is 75. Which includes those taking all three. So, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60. The number taking both physics and calculus is 30, which includes those taking all three, so the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Now, let's check if the total number of students taking each subject is consistent.

For calculus: The total number taking calculus is 180. This includes:

- Those taking only calculus.

- Those taking calculus and physics but not chemistry: 15.

- Those taking calculus and chemistry but not physics: 60.

- Those taking all three: 15.

So, total taking calculus is 15 (only calculus) + 15 (calculus and physics) + 60 (calculus and chemistry) + 15 (all three) = 15 + 15 + 60 + 15 = 105? Wait, but that's supposed to be 180. Wait, that can't be. There's a problem here. Wait, so according to this, the total number taking calculus would be 105, but we know it's 180. So, clearly, something is wrong here.

Wait, so my mistake must be in how I calculated the numbers. Let me think again.

Wait, the formula for the total number taking calculus is:

Number taking only calculus + number taking calculus and physics only + number taking calculus and chemistry only + number taking all three.

But according to the problem, the number taking both calculus and physics is 30, which includes those taking all three. So, the number taking calculus and physics only is 30 - 15 = 15. Similarly, the number taking calculus and chemistry is 75, which includes those taking all three, so the number taking calculus and chemistry only is 75 - 15 = 60. Then, the number taking all three is 15. So, the total number taking calculus is:

Only calculus + 15 (calculus and physics only) + 60 (calculus and chemistry only) + 15 (all three) = ?

But we know that total taking calculus is 180. Therefore:

Only calculus + 15 + 60 + 15 = Only calculus + 90 = 180 => Only calculus = 90.

So, the number taking only calculus is 90.

Similarly, let's check for physics. The total number taking physics is P = 110. Which includes:

Only physics + physics and calculus only + physics and chemistry only + all three.

Which is:

Only physics + 15 (physics and calculus only) + 60 (physics and chemistry only) + 15 (all three) = Only physics + 15 + 60 + 15 = Only physics + 90 = 110 => Only physics = 20.

Similarly, for chemistry, total number taking chemistry is Ch = 220. Which includes:

Only chemistry + calculus and chemistry only + physics and chemistry only + all three.

Which is:

Only chemistry + 60 (calculus and chemistry only) + 60 (physics and chemistry only) + 15 (all three) = Only chemistry + 60 + 60 + 15 = Only chemistry + 135 = 220 => Only chemistry = 85.

Now, let's check the total number of students taking at least one subject:

Only calculus (90) + Only physics (20) + Only chemistry (85) + calculus and physics only (15) + calculus and chemistry only (60) + physics and chemistry only (60) + all three (15) =

Let me add them up:

90 + 20 = 110

110 + 85 = 195

195 + 15 = 210

210 + 60 = 270

270 + 60 = 330

330 + 15 = 345.

Which matches the total of 345. So that seems okay.

But then, the problem is that when I first thought about the total number taking calculus, I thought that the numbers didn't add up, but actually, there is a "only calculus" component which is 90. So, that's okay.

So, according to this, the number of students taking physics is 110. But let me check if there are any other constraints or if I made a wrong assumption.

Wait, the problem says "Twice as many students take chemistry as take physics." So, Ch = 2P. Since we found P = 110, then Ch = 220, which matches our calculation.

Also, the number taking both calculus and chemistry is 75, which is given. The number taking both physics and chemistry is 75, which is given. The number taking both physics and calculus is 30, which is given. The number taking all three is 15.

So, all the numbers check out. Therefore, the answer should be 110.

But let me just think again if there's another way to approach this problem, maybe using a Venn diagram or another method, to confirm.

Alternatively, let's use variables for the different regions in the Venn diagram. Let me define:

Let a = only calculus

b = only physics

c = only chemistry

d = calculus and physics only

e = calculus and chemistry only

f = physics and chemistry only

g = all three

We know that:

Total students: 360

Students taking none: 15

Therefore, students taking at least one: 345 = a + b + c + d + e + f + g

Given:

g = 15

Number taking calculus: a + d + e + g = 180

Number taking physics: b + d + f + g = P

Number taking chemistry: c + e + f + g = Ch = 2P

Number taking both calculus and chemistry: e + g = 75 => e = 75 - g = 75 - 15 = 60

Number taking both physics and chemistry: f + g = 75 => f = 75 - 15 = 60

Number taking both physics and calculus: d + g = 30 => d = 30 - 15 = 15

So, we have:

d = 15, e = 60, f = 60, g = 15

Now, let's plug into the equation for calculus:

a + d + e + g = a + 15 + 60 + 15 = a + 90 = 180 => a = 90

Similarly, for chemistry:

c + e + f + g = c + 60 + 60 + 15 = c + 135 = 2P => c = 2P - 135

For physics:

b + d + f + g = b + 15 + 60 + 15 = b + 90 = P => b = P - 90

Now, total students taking at least one:

a + b + c + d + e + f + g = 90 + (P - 90) + (2P - 135) + 15 + 60 + 60 + 15

Let me compute this:

First, expand the terms:

90 + P - 90 + 2P - 135 + 15 + 60 + 60 + 15

Simplify step by step:

Start with 90 - 90 = 0

Then, 0 + P + 2P = 3P

Then, 3P - 135 + 15 + 60 + 60 + 15

Compute constants:

-135 + 15 = -120

-120 + 60 = -60

-60 + 60 = 0

0 + 15 = 15

So overall, total is 3P + 15

But we know that total is 345, so:

3P + 15 = 345 => 3P = 330 => P = 110

So, same result. Therefore, the number of students taking physics is 110.

Therefore, the answer is 110.

But let me check once more if there's any inconsistency. For example, the number taking only chemistry is c = 2P - 135 = 2*110 - 135 = 220 - 135 = 85. Which is positive, so that's okay. The number taking only physics is b = P - 90 = 110 - 90 = 20. Also positive.

So, all numbers are positive, which is good.

Therefore, I think the answer is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine how many students take **physics**, we are given a set of constraints involving three subjects: **calculus**, **physics**, and **chemistry**, and the total number of students in the school. Let's break this down step by step using the **principle of inclusion-exclusion** and logical deductions.

---

### **Step 1: Total Students and Those Taking No Subjects**

- Total students in the school: **360**
- Students who take **none** of the three subjects: **15**
- Therefore, students taking **at least one** of the subjects:
  $$
  360 - 15 = 345
  $$

---

### **Step 2: Define Variables and Known Values**

Let:
- $ C $ = number of students taking **calculus** = **180**
- $ P $ = number of students taking **physics** = **unknown**
- $ Ch $ = number of students taking **chemistry** = **unknown**
- $ |C \cap P| $ = number taking **both calculus and physics** = **30**
- $ |C \cap Ch| $ = number taking **both calculus and chemistry** = **75**
- $ |P \cap Ch| $ = number taking **both physics and chemistry** = **75**
- $ |C \cap P \cap Ch| $ = number taking **all three** = **15**

Also, it's given that:
$$
Ch = 2P
$$

---

### **Step 3: Apply Inclusion-Exclusion Principle**

The formula for the number of students taking **at least one** of the three subjects is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

$$
345 = 180 + P + Ch - 30 - 75 - 75 + 15
$$

Simplify the constants:

$$
345 = 180 + P + Ch - 180 + 15 = P + Ch + 15
$$

So:

$$
P + Ch = 330
$$

---

### **Step 4: Use the Given Relationship Between Chemistry and Physics**

We are told that:

$$
Ch = 2P
$$

Substitute into the equation:

$$
P + 2P = 330 \Rightarrow 3P = 330 \Rightarrow P = 110
$$

---

### **Step 5: Verify the Solution**

- $ P = 110 $
- $ Ch = 2P = 220 $
- Check the inclusion-exclusion formula again:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This matches the total number of students taking at least one subject, confirming the solution is consistent.

---

### **Final Answer**

$$
\boxed{110}
$$

Solution:
Let $x$ be the number of students taking physics, so the number in chemistry is $2x$.  There are 15 students taking all three, and 30 students in both physics and calculus, meaning there are $30 - 15 = 15$ students in just physics and calculus.  Similarly there are $60$ students in just chemistry and calculus, and $60$ in physics and chemistry.  Since there are $x$ students in physics and $15 + 15 + 60 = 90$ students taking physics along with other classes, $x - 90$ students are just taking physics.  Similarly, there are $2x - 135$ students taking just chemistry and $90$ students taking just calculus.  Knowing that there are 15 students not taking any of them, the sum of these eight categories is 360, the total number of people at the school:   \[
(x - 90) + (2x - 135) + 90 + 60 + 15 + 60 + 15 + 15 = 360.
\] We solve for $x$ and find that the number of physics students is $x = \boxed{110}$.


Explanation:
Okay, so Sam is carrying a 2-gallon bucket of milk, and he needs to walk 3 miles from the barn to his house. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, I need to figure out how much milk is left when he gets home. Let me think about this step by step.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. For each mile he walks, the amount of milk decreases to 2/3 of what it was at the beginning of that mile. So, does that mean that after each mile, the milk is multiplied by 2/3? Or is there some other way the leak is affecting the milk?

Let me parse the wording again: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, when he starts walking the first mile, the bucket has 2 gallons. After walking that mile, the amount of milk is 2/3 of the original amount. So, after the first mile, it's 2*(2/3) = 4/3 gallons? Wait, but that seems like the milk is decreasing by a factor of 2/3 each mile. So, each mile, the milk is multiplied by 2/3. Therefore, after each mile, the amount is (2/3) times the previous amount.

So, if that's the case, then after the first mile, it's 2*(2/3). After the second mile, it's 2*(2/3)^2. After the third mile, it's 2*(2/3)^3. Therefore, since he walks 3 miles, the total milk left would be 2*(2/3)^3. Let me check if that makes sense.

Wait, but maybe I need to think about it differently. Maybe the leak is causing the milk to decrease continuously as he walks, not just at the end of each mile? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, the wording suggests that after completing each mile, the amount is 2/3 of what it was at the beginning of that mile. So, it's a discrete decrease every mile, not a continuous leak. So, each mile, the amount is multiplied by 2/3. Therefore, after 3 miles, it's 2*(2/3)^3.

Let me compute that. 2*(2/3)^3. First, (2/3)^3 is 8/27. Then 2*(8/27) is 16/27. So, 16/27 gallons. Is that the answer? Wait, but let me make sure I didn't misinterpret the problem.

Alternatively, maybe the problem is that for each mile, the amount of milk decreases by 2/3? But that would be different. If it's decreasing by 2/3, then after the first mile, it would be 2 - (2/3)*2 = 2 - 4/3 = 2/3. But the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that would mean that the amount is 2/3 of the original, not that it's decreased by 2/3. So, the wording is important here. If it's "2/3 as much," that is, the amount is multiplied by 2/3. So, the first interpretation is correct.

But let me think again. Suppose he walks one mile. At the beginning of the mile, he has 2 gallons. Then, after walking that mile, the amount is 2/3 of that, so 4/3 gallons. Then, he starts the second mile with 4/3 gallons. After walking the second mile, it's 2/3 of 4/3, which is 8/9. Then, after the third mile, it's 2/3 of 8/9, which is 16/27. So, yes, that seems to be the case. Therefore, the answer is 16/27 gallons. So, 16 divided by 27 is approximately 0.592 gallons. But since the question asks for the exact value, 16/27 is the answer.

But let me check if there's another way to interpret the problem. Maybe the leak is such that for each mile walked, the milk decreases by 2/3 of the original amount? But that would be different. For example, if he walks one mile, he loses 2/3 of the original 2 gallons, so 4/3 gallons, leaving 2 - 4/3 = 2/3 gallons. Then, for the next mile, he would lose 2/3 of the original 2 gallons again? But that seems odd because the amount of milk is decreasing each mile. If he loses 2/3 of the original amount each mile, then after the first mile, he has 2 - 2*(2/3) = 2 - 4/3 = 2/3. Then, for the second mile, he would lose 2/3 of the original 2 gallons again, so 4/3, but he only has 2/3 left. That would mean he would have negative milk, which doesn't make sense. Therefore, that interpretation is probably wrong.

Alternatively, maybe the leak is such that for each mile, the milk decreases by 2/3 of the current amount? That is, each mile, the milk is reduced by 2/3 of what it currently is. So, that would be a different scenario. For example, starting with 2 gallons. After the first mile, he loses 2/3 of 2 gallons, which is 4/3, so remaining is 2 - 4/3 = 2/3. Then, for the second mile, he loses 2/3 of the current amount, which is 2/3 * 2/3 = 4/9, so remaining is 2/3 - 4/9 = 2/9. Then, third mile, he loses 2/3 of 2/9, which is 4/27, so remaining is 2/9 - 4/27 = 2/27. So, that would be 2/27 gallons. But this is different from the previous answer. However, the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, this wording suggests that after walking the mile, the amount is 2/3 of the beginning amount. So, if you start with X, after the mile, it's (2/3)X. Therefore, the first interpretation is correct. So, the answer would be 2*(2/3)^3 = 16/27. However, the second interpretation, where each mile you lose 2/3 of the current amount, would be a different problem. But the wording is crucial here.

Let me check the exact wording again: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, when he starts walking the first mile, the amount is 2 gallons. After walking that mile, the amount is 2/3 of the beginning of the mile. So, that is, after the first mile, it's 2*(2/3). Then, when he starts the second mile, the amount is 2*(2/3). After walking the second mile, it's 2*(2/3)*(2/3). Similarly, after the third mile, it's 2*(2/3)^3. Therefore, the answer is 16/27. Therefore, I think that is the correct answer.

But let me think again. Suppose that the leak is such that during each mile, the milk decreases to 2/3 of what it was at the beginning of the mile. So, the decrease happens over the mile, but the amount at the end of the mile is 2/3 of the beginning. So, that would mean that the rate of leakage is such that after one mile, it's 2/3. So, that would be a multiplicative factor each mile. Therefore, the answer is 16/27. Therefore, I think that is correct.

Alternatively, if the problem had said that for each mile, the milk decreases by 2/3 of the original amount, then it would be different, but that's not what it says. It says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, the amount at the end of the mile is 2/3 of the beginning. Therefore, the answer is 16/27.

But let me check with another approach. Let's think of it as a geometric sequence. The initial amount is 2 gallons. After each mile, the amount is multiplied by 2/3. Therefore, after 3 miles, the amount is 2*(2/3)^3. Which is 2*(8/27) = 16/27. So, that's the same as before.

Alternatively, if we model it as a continuous process, but the problem says "for each mile he walks," which suggests that the decrease is per mile, not continuously. Therefore, the answer is 16/27.

But let me check if there's a different interpretation. Suppose that the leak is such that for each mile walked, the milk decreases by a factor of 2/3. So, the amount of milk is multiplied by 2/3 each mile. So, same as before. So, same answer.

Alternatively, maybe the problem is that the bucket leaks 2/3 of a gallon per mile? But that would be different. If it's leaking 2/3 gallon per mile, then after 3 miles, he would have lost 3*(2/3) = 2 gallons, so he would have 0 gallons left. But that contradicts the initial amount. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that wording is not about the amount lost, but about the remaining amount. So, the remaining amount is 2/3 of the beginning amount. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the rate of leakage is such that the milk decreases exponentially. For example, if the leak is proportional to the amount of milk, then the amount would decrease exponentially. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that is a discrete decrease, not a continuous one. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the leak causes the milk to decrease by 2/3 every mile, meaning that after each mile, the amount is 1 - 2/3 = 1/3 of the original? But that would be a different interpretation. But the wording says "there is 2/3 as much milk," so that would be 2/3 of the original. So, that would be the same as multiplying by 2/3.

Therefore, I think the answer is 16/27. Let me check with another example. Suppose he walks 1 mile. Then, the milk is 2*(2/3) = 4/3. If he walks another mile, it's 4/3*(2/3) = 8/9. Then, third mile, 8/9*(2/3) = 16/27. So, yes, that seems correct.

Alternatively, if the problem had said that the milk is reduced by 2/3 each mile, meaning that after each mile, the amount is 1/3 of the original, then the answer would be 2*(1/3)^3 = 2/27. But that's not the case here.

Therefore, I think the answer is 16/27. Therefore, the number of gallons of milk when Sam gets home is 16/27.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam's bucket when he reaches his house, we analyze the situation step by step.

---

### **Understanding the Problem**

Sam starts with a **2-gallon bucket of milk** and walks **3 miles** to his house. However, the bucket has a **leak**, and the problem states:

> *"For each mile he walks, there is $\frac{2}{3}$ as much milk in the bucket as at the beginning of the mile."*

This means that **after each mile**, the amount of milk in the bucket is **multiplied by $\frac{2}{3}$** of what it was at the **start of that mile**.

---

### **Modeling the Situation**

Let’s denote the amount of milk at the **start of the $n$-th mile** as $M_n$. Then, the amount of milk at the **end of the $n$-th mile** is:

$$
M_{n+1} = M_n \times \frac{2}{3}
$$

This is a **geometric sequence** where each term is $\frac{2}{3}$ of the previous one.

Starting with $M_0 = 2$ gallons (the initial amount), we can compute the amount of milk after each mile:

- After 1st mile: $M_1 = 2 \times \frac{2}{3} = \frac{4}{3}$
- After 2nd mile: $M_2 = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$
- After 3rd mile: $M_3 = \frac{8}{9} \times \frac{2}{3} = \frac{16}{27}$

---

### **Final Calculation**

So, after walking **3 miles**, the amount of milk remaining in the bucket is:

$$
M_3 = 2 \times \left(\frac{2}{3}\right)^3 = 2 \times \frac{8}{27} = \frac{16}{27}
$$

---

### **Conclusion**

Therefore, the amount of milk in the bucket when Sam reaches his house is:

$$
\boxed{\dfrac{16}{27}}
$$

[Assistant]
Explanation:
Okay, so Sam is carrying a 2-gallon bucket of milk, and he needs to walk 3 miles from the barn to his house. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, I need to figure out how much milk is left when he gets home. Let me think about this step by step.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. For each mile he walks, the amount of milk decreases to 2/3 of what it was at the beginning of that mile. So, does that mean that after each mile, the milk is multiplied by 2/3? Or is there some other way the leak is affecting the milk?

Let me parse the wording again: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, when he starts walking the first mile, the bucket has 2 gallons. After walking that mile, the amount of milk is 2/3 of the original amount. So, after the first mile, it's 2*(2/3) = 4/3 gallons? Wait, but that seems like the milk is decreasing by a factor of 2/3 each mile. So, each mile, the milk is multiplied by 2/3. Therefore, after each mile, the amount is (2/3) times the previous amount.

So, if that's the case, then after the first mile, it's 2*(2/3). After the second mile, it's 2*(2/3)^2. After the third mile, it's 2*(2/3)^3. Therefore, since he walks 3 miles, the total milk left would be 2*(2/3)^3. Let me check if that makes sense.

Wait, but maybe I need to think about it differently. Maybe the leak is causing the milk to decrease continuously as he walks, not just at the end of each mile? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, the wording suggests that after completing each mile, the amount is 2/3 of what it was at the beginning of that mile. So, it's a discrete decrease every mile, not a continuous leak. So, each mile, the amount is multiplied by 2/3. Therefore, after 3 miles, it's 2*(2/3)^3.

Let me compute that. 2*(2/3)^3. First, (2/3)^3 is 8/27. Then 2*(8/27) is 16/27. So, 16/27 gallons. Is that the answer? Wait, but let me make sure I didn't misinterpret the problem.

Alternatively, maybe the problem is that for each mile, the amount of milk decreases by 2/3? But that would be different. If it's decreasing by 2/3, then after the first mile, it would be 2 - (2/3)*2 = 2 - 4/3 = 2/3. But the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that would mean that the amount is 2/3 of the original, not that it's decreased by 2/3. So, the wording is important here. If it's "2/3 as much," that is, the amount is multiplied by 2/3. So, the first interpretation is correct.

But let me think again. Suppose he walks one mile. At the beginning of the mile, he has 2 gallons. Then, after walking that mile, the amount is 2/3 of that, so 4/3 gallons. Then, he starts the second mile with 4/3 gallons. After walking the second mile, it's 2/3 of 4/3, which is 8/9. Then, after the third mile, it's 2/3 of 8/9, which is 16/27. So, yes, that seems to be the case. Therefore, the answer is 16/27 gallons. So, 16 divided by 27 is approximately 0.592 gallons. But since the question asks for the exact value, 16/27 is the answer.

But let me check if there's another way to interpret the problem. Maybe the leak is such that for each mile walked, the milk decreases by 2/3 of the original amount? But that would be different. For example, if he walks one mile, he loses 2/3 of the original 2 gallons, so 4/3 gallons, leaving 2 - 4/3 = 2/3 gallons. Then, for the next mile, he would lose 2/3 of the original 2 gallons again? But that seems odd because the amount of milk is decreasing each mile. If he loses 2/3 of the original amount each mile, then after the first mile, he has 2 - 2*(2/3) = 2 - 4/3 = 2/3. Then, for the second mile, he would lose 2/3 of the original 2 gallons again, so 4/3, but he only has 2/3 left. That would mean he would have negative milk, which doesn't make sense. Therefore, that interpretation is probably wrong.

Alternatively, maybe the leak is such that for each mile, the milk decreases by 2/3 of the current amount? That is, each mile, the milk is reduced by 2/3 of what it currently is. So, that would be a different scenario. For example, starting with 2 gallons. After the first mile, he loses 2/3 of 2 gallons, which is 4/3, so remaining is 2 - 4/3 = 2/3. Then, for the second mile, he loses 2/3 of the current amount, which is 2/3 * 2/3 = 4/9, so remaining is 2/3 - 4/9 = 2/9. Then, third mile, he loses 2/3 of 2/9, which is 4/27, so remaining is 2/9 - 4/27 = 2/27. So, that would be 2/27 gallons. But this is different from the previous answer. However, the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, this wording suggests that after walking the mile, the amount is 2/3 of the beginning amount. So, if you start with X, after the mile, it's (2/3)X. Therefore, the first interpretation is correct. So, the answer would be 2*(2/3)^3 = 16/27. However, the second interpretation, where each mile you lose 2/3 of the current amount, would be a different problem. But the wording is crucial here.

Let me check the exact wording again: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, when he starts walking the first mile, the amount is 2 gallons. After walking that mile, the amount is 2/3 of the beginning of the mile. So, that is, after the first mile, it's 2*(2/3). Then, when he starts the second mile, the amount is 2*(2/3). After walking the second mile, it's 2*(2/3)*(2/3). Similarly, after the third mile, it's 2*(2/3)^3. Therefore, the answer is 16/27. Therefore, I think that is the correct answer.

But let me think again. Suppose that the leak is such that during each mile, the milk decreases to 2/3 of what it was at the beginning of the mile. So, the decrease happens over the mile, but the amount at the end of the mile is 2/3 of the beginning. So, that would mean that the rate of leakage is such that after one mile, it's 2/3. So, that would be a multiplicative factor each mile. Therefore, the answer is 16/27. Therefore, I think that is correct.

Alternatively, if the problem had said that for each mile, the milk decreases by 2/3 of the original amount, then it would be different, but that's not what it says. It says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, the amount at the end of the mile is 2/3 of the beginning. Therefore, the answer is 16/27.

But let me check with another approach. Let's think of it as a geometric sequence. The initial amount is 2 gallons. After each mile, the amount is multiplied by 2/3. Therefore, after 3 miles, the amount is 2*(2/3)^3. Which is 2*(8/27) = 16/27. So, that's the same as before.

Alternatively, if we model it as a continuous process, but the problem says "for each mile he walks," which suggests that the decrease is per mile, not continuously. Therefore, the answer is 16/27.

But let me check if there's a different interpretation. Suppose that the leak is such that for each mile walked, the milk decreases by a factor of 2/3. So, the amount of milk is multiplied by 2/3 each mile. So, same as before. So, same answer.

Alternatively, maybe the problem is that the bucket leaks 2/3 of a gallon per mile? But that would be different. If it's leaking 2/3 gallon per mile, then after 3 miles, he would have lost 3*(2/3) = 2 gallons, so he would have 0 gallons left. But that contradicts the initial amount. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that wording is not about the amount lost, but about the remaining amount. So, the remaining amount is 2/3 of the beginning amount. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the rate of leakage is such that the milk decreases exponentially. For example, if the leak is proportional to the amount of milk, then the amount would decrease exponentially. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that is a discrete decrease, not a continuous one. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the leak causes the milk to decrease by 2/3 every mile, meaning that after each mile, the amount is 1 - 2/3 = 1/3 of the original? But that would be a different interpretation. But the wording says "there is 2/3 as much milk," so that would be 2/3 of the original. So, that would be the same as multiplying by 2/3.

Therefore, I think the answer is 16/27. Let me check with another example. Suppose he walks 1 mile. Then, the milk is 2*(2/3) = 4/3. If he walks another mile, it's 4/3*(2/3) = 8/9. Then, third mile, 8/9*(2/3) = 16/27. So, yes, that seems correct.

Alternatively, if the problem had said that the milk is reduced by 2/3 each mile, meaning that after each mile, the amount is 1/3 of the original, then the answer would be 2*(1/3)^3 = 2/27. But that's not the case here.

Therefore, I think the answer is 16/27. Therefore, the number of gallons of milk when Sam gets home is 16/27.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam's bucket when he reaches his house, we analyze the situation step by step.

---

### **Understanding the Problem**

Sam starts with a **2-gallon bucket of milk** and walks **3 miles** to his house. However, the bucket has a **leak**, and the problem states:

> *"For each mile he walks, there is $\frac{2}{3}$ as much milk in the bucket as at the beginning of the mile."*

This means that **after each mile**, the amount of milk in the bucket is **multiplied by $\frac{2}{3}$** of what it was at the **start of that mile**.

---

### **Modeling the Situation**

Let’s denote the amount of milk at the **start of the $n$-th mile** as $M_n$. Then, the amount of milk at the **end of the $n$-th mile** is:

$$
M_{n+1} = M_n \times \frac{2}{3}
$$

This is a **geometric sequence** where each term is $\frac{2}{3}$ of the previous one.

Starting with $M_0 = 2$ gallons (the initial amount), we can compute the amount of milk after each mile:

- After 1st mile: $M_1 = 2 \times \frac{2}{3} = \frac{4}{3}$
- After 2nd mile: $M_2 = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$
- After 3rd mile: $M_3 = \frac{8}{9} \times \frac{2}{3} = \frac{16}{27}$

---

### **Final Calculation**

So, after walking **3 miles**, the amount of milk remaining in the bucket is:

$$
M_3 = 2 \times \left(\frac{2}{3}\right)^3 = 2 \times \frac{8}{27} = \frac{16}{27}
$$

---

### **Conclusion**

Therefore, the amount of milk in the bucket when Sam reaches his house is:

$$
\boxed{\dfrac{16}{27}}
$$

Solution: At the end of the first mile, there will be $\frac{2}{3}$ of the initial milk in the bucket. Each additional mile multiplies this amount by $\frac{2}{3}$. Thus, when he arrives at home at the end of the third mile, there will be $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \left(\frac{2}{3}\right)^{3}$ as much milk in the bucket. Since he had 2 gallons initially, the amount in the bucket when he gets home is $2 \cdot \left(\frac{2}{3}\right)^{3}$. Because $\left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}$, this expression is equivalent to $2 \cdot \frac{2^{3}}{3^{3}}$. Because $n^{a} \cdot n^{b} = n^{a+b}$, this equals $\frac{2^{4}}{3^{3}}$. Multiplying the exponents out, we get $\boxed{\frac{16}{27}}$ gallons.