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?

  • $\begingroup$ For the second question, you would only subtract out those committees with no sophomores or one sophomore. $\endgroup$ Commented Feb 2, 2018 at 10:33

2 Answers 2


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 ${28 \choose 8} - {10 \choose 0} * {18 \choose 8} - {10 \choose 1} * {18 \choose 7} = 2746107.$


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 $2!$.

  • $\begingroup$ Your second answer is incorrect. You correctly selected six people in the first term. However, you selected eight people in the second and third terms. Also, in the first answer, $\binom{28}{6} = \frac{28!}{6!22!}$. The factor of $6!$ in the denominator of the simplified form you wrote represents the number of orders in which the same six people be selected. $\endgroup$ Commented Mar 30, 2018 at 13:46
  • $\begingroup$ @N.F.Taussig: Thank you so much... I meant $8$, but I just copied and paste from the first answer... $\endgroup$ Commented Mar 30, 2018 at 14:38
  • $\begingroup$ And apparently my first explanation is wrong, sorry I'm editing it now $\endgroup$ Commented Mar 30, 2018 at 14:45
  • $\begingroup$ Your answer is now correct. Another way of thinking about the second problem is this: The number of ways we can select $k$ of the $10$ sophomores and $8 - k$ of the remaining $12 + 6$ students is $\binom{8}{k}\binom{12 + 6}{8 - k}$, so the number of ways of selecting at least two sophomores is $$\binom{28}{8} - \binom{10}{0}\binom{12 + 6}{8} - \binom{10}{1}\binom{12 + 6}{7}$$ which is equivalent to your answer. $\endgroup$ Commented Mar 30, 2018 at 14:52

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