<Question>
Mr. Brennan has 7 boys and 4 girls in his statistics class. In how many ways can he pick 3 boys and 2 girls to give a group presentation tomorrow? (The order in which the boys and girls are chosen does not matter.)
</Question>

<think>
<Plan_1>
Okay, so Mr. Brennan has a total of 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 different ways he can do this, and the order in which the boys and girls are chosen doesn't matter. Hmm, let's see. I think this is a combinatorics problem, right? Because we're dealing with combinations where the order doesn't matter.

First, let me recall the basics of combinations. The formula for combinations is C(n, k) = n! / (k! * (n - k)! ), where n is the total number of items, k is the number we want to choose, and ""! denotes factorial. So, combinations are used here because the problem specifies that the order doesn't matter. For example, choosing Boy A and Boy B is the same as choosing Boy B and Boy A in the group, so we don't want to count those as separate groups.

Since Mr. Brennan needs to pick 3 boys out of 7 and 2 girls out of 4, we can handle these two selections separately and then multiply the results.
That makes sense because for each combination of boys, there are multiple combinations of girls that can be paired with them. So, the total number of ways is the product of the number of ways to choose the boys and the number of ways to choose the girls.

Let me start with the boys. There are 7 boys, and we need to choose 3. Using the combination formula, that would be C(7, 3). Let me compute that.

<prompt_1.1> Calculate the number of ways to choose 3 boys out of 7 using the combination formula. </prompt_1.1>

Now, moving on to the girls. There are 4 girls, and we need to choose 2. So that's C(4, 2).
Calculating that: 4! / (2! * (4 - 2)! ) = (4 * 3 * 2 * 1) / (2 * 1 * 2 * 1) = 24 / 4 = 6. So there are 6 ways to choose 2 girls out of 4.
-----

Let me just check if there's another way to approach this problem. Maybe using permutations? But no, since the order doesn't matter, permutations would involve order, so combinations are the right approach here.

Another possible mistake could be mixing up combinations and permutations. For instance, if someone thought order mattered, they might use permutations instead. Let's see what happens if they did that.

<prompt_1.2> Discuss the importance of using combinations over permutations for this problem. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
7! / (3! * (7 - 3)! ) = (7 * 6 * 5 * 4 * 3 * 2 * 1) / (6 * 3 * 2 * 1) Wait, actually, simplifying that, since 7! is 7*6*5*4! and the denominator is 3! * 4! . So, cancel out the 4! terms. That leaves (7*6*5)/(3*2*1). Calculating that: 7*6 is 42, 42*5 is 210. Then, 3*2*1 is 6. So 210 divided by 6 is 35. So there are 35 ways to choose the boys.
</execution_1.1>

<execution_1.2>
If they used permutations for both, the calculation would be P(7,3) * P(4,2). But P(n,k) = n! / (n - k)! , so P(7,3) = 7*6*5 = 210, and P(4,2) = 4*3 = 12. Then 210 * 12 = 2520, which is way higher. But since the problem says order doesn't matter, that's not the case here. So definitely combinations.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Then, as established, the girls' combinations are 6. So the total number of groups is 35 * 6. Let me calculate that.
35 multiplied by 6 is 210. So, 210 different ways.

Wait a second, let me verify that again.
If there are 35 ways to choose the boys and 6 ways to choose the girls, then for each of the 35 boy groups, there are 6 possible girl groups that can be paired with them. So, yes, 35 * 6 = 210 total groups.
-----

Based on execution_1.1, execution_1.2:
Let me just verify the combinations again.

<prompt_2.1> Verify the calculations for combinations of boys and girls. </prompt_2.1>
-----

Another check: total number of students is 7 + 4 = 11. If order mattered for the entire group, it would be 11! / ( (11 -5)! ), but since we are splitting into boys and girls, it's better to handle them separately.

<prompt_2.2> Consider the overall group selection and confirm the approach of separating boys and girls. </prompt_2.2>

Let me just make sure there isn't any overlap or constraints I missed.
The problem states 7 boys and 4 girls, no mixed selection, just 3 boys and 2 girls. So, no issues there.
-----

Therefore, I think the answer is 210. But just to be thorough, let me compute C(7,3) and C(4,2) again.

<prompt_2.3> Re-compute and verify the combination calculations for boys and girls. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
C(7,3) = 35, as 7*6*5 / 3*2*1 = 35. Yes. C(4,2) = 6, as 4*3 / 2*1 = 6. So 35*6=210.
</execution_2.1>

<execution_2.2>
But since the group is 5 people, 3 boys and 2 girls, the total number of ways would be C(7,3) * C(4,2). Which is exactly what we did. So that's the right approach.

Alternatively, if we tried to compute the total number of ways to choose 5 students out of 11, that would be C(11,5). But that's not considering the gender split. So in this problem, the gender split is important, so we have to handle the boys and girls separately and then combine the results.
</execution_2.2>

<execution_2.3>
C(7,3): 7 choose 3. As above, 7*6*5 / 3*2*1 = 35. Correct. C(4,2): 4*3 / 2*1 = 6. Correct. 35*6=210.
</execution_2.3>
</Execution_2>

<Final_answer>
Therefore, the number of ways Mr. Brennan can pick 3 boys and 2 girls from his class for the group presentation is $\boxed{210}$.
</Final_answer>
</think>