Mathematics Asked by Scott on December 16, 2020

Given 10 people $P_1, P_2, ldots,P_{10}$ how many 6-member teams can be formed if at most one of $P_2, P_4$ can be chosen?

Either one of P2 or P4 gives 8c5 ways to choose.

My answer is:

$$

{8choose5} + {8choose5} + {8choose6} = 140

$$

Is my answer correct?

Your answer is correct.

Another method of doing the problem is to subtract those selections in which both $P_2$ and $P_4$ are selected from the total number of $6$-member teams which could be formed from the $10$ people. There are $binom{10}{6}$ ways to select six of the ten people. If both $P_2$ and $P_4$ were selected, we would have to select four of the other $10 - 2 = 8$ people. Hence, the number of $6$-members teams which include at most one of $P_2$ and $P_4$ is $$binom{10}{6} - binom{2}{2}binom{8}{4} = 210 - 70 = 140$$ as you found by a direct count.

Correct answer by N. F. Taussig on December 16, 2020

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