# Given 10 people: $P_1, P_2, ldots,P_{10}$ how many 6-member teams can be formed if

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

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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