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A school is forming a group of 5. There are 10 freshman and 6 seniors, and the group must have at least 2 freshman and at least 1 senior. How many distinct groups are possible?

My approach (that I think is definitely flawed) -

This would be a combination problem because order does not matter. There has to be at least 2 freshmen and 1 senior so there are only 2 open spots left with a pool of 8 freshmen and 5 seniors, or 13 people. Would it just be C(13,2) or 13!/(2*11!)?

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  • $\begingroup$ You should accept one of the questions provided by Dylan and Arnold Frenzy as an answer to your question :) This will confirm that you have found a satisfying answer and they will earn reputation for answering your question. $\endgroup$ – derthomas Jan 31 '17 at 2:57
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While your approach is definitely flawed, you did get to the root of the problem!

You choose the remaining spots perfectly well, but you do not consider how many possible choices there are for the first 3 spots. So, while "order does not matter" (i.e. you will be using combinations) there are multiple combinations of individuals to choose. As a matter of fact, we have $10$ freshman and must choose $2$, and we have $6$ seniors and must choose one.

The temptation is to, at this point, simply take ${10 \choose 2}{6 \choose 1}{13 \choose 2}$. But this won't work, as we are now going to be double counting, that is, when we have the $13 \choose 2$ those will be considerations that we have already made in the previously selections.

As such, there are a couple ways to approach such a counting problem. One way, that is less elegant, but definitely instructive, is to enumerate the possible counting combinations. That is we could have:

\begin{align} \begin{array}{c | c | c} Freshman & Seniors & Combinations \\ \hline 4 & 1 & {10 \choose 4}{6 \choose 1} = 1260\\ 3 & 2 & {10 \choose 3}{6 \choose 2} = 1800 \\ 2 & 3 & {10 \choose 2}{6 \choose 3} = 900 \\ \hline & & Total = 3960 \end{array} \end{align}

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As you say the order doesn't matter. Then you have to think of all allowed groups, that is 2 freshmen 3 seniors, 3 freshmen 2 seniors, and 4 freshmen 1 senior.

2 freshmen 3 seniors:

$\dfrac{10 \cdot 9}{2!} \cdot \dfrac{6 \cdot 5 \cdot 4}{3!} = \dfrac{10!}{2!(10-2)!}\cdot \dfrac{6!}{3!(6-3)!} = 900 $ possible combinations

3 freshmen 2 seniors:

$\dfrac{10 \cdot 9 \cdot 8}{3!} \cdot \dfrac{6 \cdot 5}{2!} = \dfrac{10!}{3!(10-3)!}\cdot \dfrac{6!}{2!(6-2)!} = 1,800 $ possible combinations

4 freshmen 1 senior:

$\dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4!} \cdot 6 = \dfrac{10!}{(10-4)!4!}\cdot \dfrac{6!}{1!(6-1)!} = 1,260 $ possible combinations

TOTAL:

Therefore, in total, there are $3,960$ possibilities

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Other users have already given the right answer, but I would like to add why your approach does not count all the possibilities. Let's try to understand the error in the reasoning by considering a somewhat simpler example where you have two seniors, denoted by $S_1$ and $S_2$ and two freshmen denoted by $F_1$ and $F_2$ and you form groups of 2 people where at least one of them has to be a senior.

If we follow your approach, then we would have only 3 possibilities, since we leave one spot open for the set that contains one senior and 2 freshmen after having filled the one obligatory spot with a senior.

Suppose that $S_1$ is the senior that fills the obligatory senior spot. Then we have the following possible groups: $S_1S_2$, $S_1F_1$ and $S_1F_2$. But where is $S_2F_2$ or $S_2F_1$ in this scenario? You get the idea: we have counted only half of the possible groups.

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