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Five people are to be chosen from a group of ten people, to form a committee.

(b) Amongst the ten people, we have Alice, Bob, Eve and Oscar. Alice and Bob are extremely talkative during meetings, and so at most one of these two can be chosen. Eve and Oscar are inseparable, and so a committee must either contain both of them or neither of them. With these constraints, how many different five-person committees can be formed?

I have the solution already, I just want to understand how to express 'either' and 'at most' and understand the solution. For the following answers, I don't understand the $2$ or $6$ in the ncr.

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N. F. Taussig
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AliceRathbone
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  • I don't know what you mean by "ncr." The solution is essentially counting the number of possible combinations of committee members who are not any of Alice, Bob, Eve, or Oscar. So, for example, if none of them are on the committee, you must choose $5$ members from the remaining $6$ people, which gives you $\binom 65=6$. – Robert Shore Jun 05 '23 at 16:13
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    I think "ncr" means ${_n}{C_r} = C(n,r) = {{n}\choose{r}}$. – mjqxxxx Jun 05 '23 at 16:17
  • @RobertShore Thanks Robert, can you explain the first scenario? – AliceRathbone Jun 05 '23 at 17:32
  • If Alice is included and Eve and Oscar are out, there are $\binom 64=\binom 62$ ways to choose the other committee members. (Choosing $4$ out of $6$ to include is the same as choosing $2$ out of $6$ to exclude -- that's the combinatorial explanation for why $\binom 64=\binom 62$.) It's exactly the same calculation if Bob is included and Eve and Oscar are out, so multiply that number by $2$ to account for both cases. – Robert Shore Jun 05 '23 at 23:30
  • You have been asked before not to use pictures of text. – Shaun Jun 06 '23 at 13:11

1 Answers1

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We can divide the people into three groups: Alice and Bob, Eve and Oscar, and the other six people. The number of ways of selecting $k$ people from the group that includes Alice and Bob, $m$ people from the group that includes Eve and Oscar, and $n$ people from the remaining six people is $$\binom{2}{k}\binom{2}{m}\binom{6}{n}$$

The problem requires that we select at most one person from the group consisting of Alice and Bob, either zero or two people from the group that consisting of Eve and Oscar, and the remaining $n = 5 - k - m$ people from among the remaining six people. Note that at most one means either $0$ or $1$ people since the number of people we select must be a nonnegative integer.

Let's consider cases:

One of Alice or Bob, neither Eve nor Oscar: If we select one person from the group consisting of Alice and Bob and no people from the group consisting of Eve and Oscar, we must select $5 - 1 - 0 = 4$ people from the remaining six people. There are $$\binom{2}{1}\binom{2}{0}\binom{6}{4} = 2 \cdot 1 \cdot \binom{6}{4} = 2\binom{6}{4}$$ such selections.

Neither Alice nor Bob, neither Eve nor Oscar: As Robert Shore explained in the comments, this means all five people must be selected from the group of six people, which can be done in $$\binom{2}{0}\binom{2}{0}\binom{6}{5} = 1 \cdot 1 \cdot \binom{6}{5} = \binom{6}{5}$$ ways.

One of Alice or Bob, both Eve and Oscar: If we select one person from the group consisting of Alice and Bob and select both people in the group consisting of Eve and Oscar, we must select $5 - 1 - 2 = 2$ people from among the remaining six people. There are $$\binom{2}{1}\binom{2}{2}\binom{6}{2} = 2 \cdot 1 \cdot \binom{6}{2} = 2\binom{6}{2}$$ such selections.

Neither Alice nor Bob, both Eve and Oscar: Can you finish this case on your own?

N. F. Taussig
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