Okay, so Mr. Brennan has 7 boys and 4 girls in his statistics class. He needs to pick 3 boys and 2 girls to give a group presentation. The question is asking in how many ways he can do this. Hmm, let's see. Since the order doesn't matter, this is a combination problem, right? Because combinations are used when the order isn't important, unlike permutations where order matters.

Alright, so first, I need to figure out how many ways he can choose 3 boys out of 7. Then, separately, how many ways he can choose 2 girls out of 4. Then, since these two choices are independent of each other, the total number of ways should be the product of those two numbers. That makes sense because for each combination of boys, there are multiple combinations of girls that can go with them.

Let me recall the formula for combinations. The combination formula is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. So, applying this formula to both the boys and the girls.

First, for the boys: C(7, 3). Let me compute that. 7! divided by (3! * (7 - 3)!)) which is 7! / (3! * 4!) Let's calculate this. 7! is 7*6*5*4*3*2*1, but since 4! is in the denominator, we can cancel out the 4! part. So, 7! / 4! is 7*6*5. Then divide by 3! which is 6. So 7*6*5 / 6 = 7*5 = 35. So there are 35 ways to choose the boys.

Now for the girls: C(4, 2). Applying the same formula. 4! / (2! * (4 - 2)! ) = 4! / (2! * 2! ) Let's compute that. 4*3*2*1 divided by (2*1 * 2*1) which is 24 / 4 = 6. So there are 6 ways to choose the girls.

Therefore, the total number of ways to pick 3 boys and 2 girls is 35 multiplied by 6. Let me do that multiplication: 35 * 6. 30*6 is 180, and 5*6 is 30, so 180 + 30 = 210. So the total number of ways is 210.

Wait, let me double-check my calculations to make sure I didn't make a mistake. For the boys: C(7,3). Let's compute 7 choose 3. 7*6*5 / 3*2*1. 7*6 is 42, 42*5 is 210. Divided by 6 (which is 3*2*1). 210 / 6 is 35. Correct. For the girls: 4 choose 2. 4*3 / 2*1 = 12 / 2 = 6. Correct. Then 35*6 is indeed 210. So that seems right.

Another way to think about it: For each boy, we're selecting 3 out of 7. The girls are 2 out of 4. Since these are independent selections, the total combinations are the product. So yes, 35*6=210. That's the answer.
Therefore, the number of ways Mr. Brennan can choose 3 boys and 2 girls for the presentation is $\boxed{210}$.
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