Basic combinatoric question

Question

A class contains 10 sophomores, 12 juniors, and 6 seniors.

a) How many ways can we form a group of 6 people?

I'm not sure if each student within their group should be considered as distinguishable individuals or not. If yes, would this just be ${28\choose 6}$?

b) How many ways can we form a group of 8 that contain at least 2 sophomores?

So in other words, subtract out the number of possibilities for exactly 0 sophomore, 1 sophomore, and 2 sophomores? How would one go about that and are they distinguishable?

Answer 1

If all of the students are distinct (I would hope this is the case), then for part a the answer is ${28 \choose 6}$, as you have 28 unique objects, of which you pick 6.

For part b, you are correct in your approach - you are counting the cases which do not work, and removing then from your total cases.

In a group of $y$ with $x$ sophomores, you have ${10 \choose x}$ ways to choose the sophomores, since there are 10 sophomores and we want x of them, then we have ${12+6 \choose y-x}$ ways to choose the remaining members of the group. We get $12+6$ as our total number of students to pick because the remaining members of the group cannot be sophomores; they must be either a junior or a senior, of which there are 12 and 6 of, respectively.

We then multiply these 2 quantities, ${10 \choose x}$ and ${12+6 \choose y-x}$ to get the total number of ways to pick a group of $y$ students with the requirement that $x$ of them are sophomores.

In part B we are picking a group of 8 students, with the requirement that at least 2 students are sophomores. Therefore, we want to subtract the cases where we have 0 sophomores and 1 sophomore.

So, the answer to part b is (Total Possibilities - Possibilities with 0 sophomores - Possibilities with 1 sophomore), or ${28 \choose 8} - {10 \choose 0} * {18 \choose 8} - {10 \choose 1} * {18 \choose 7} = 2746107.$

Answer 2

a) $$ \binom{28}{6}=\frac{28!}{6!22!} $$

It's people that make a group distinct, not the titles they have. So just choose $6$ out of $28$ and you don't need to divide it by anything related to the titles.

b) $$ Group\ of\ 8\ people\ -\ No\ sophomore\ -\ 1\ sophomore\\ \binom{28}{8} - \binom{12+6}{8} - 10\cdot\binom{12+6}{7}\\ $$

Considering $10\cdot\binom{12+6}{7}$, since the former is a person and the latter is a group. They're of different types, so you don't have to divide it by $2!$.