# probability question - Please Correct me

Thirteen people on a softball team show up for a game. Of the 13 people who show up three are women.How many ways are there to choose 10 players to take the field if at least two of these players must be a woman?

Please Correct Me, Total = C(13,10) if only men show up C(10,10) if one woman show up C(10,9)

ans = C(13,10) - C(10,10) - C(10,9)

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Not quite: there is $1$ way to choose a team with no women, and there are $3\binom{10}9$ ways to choose a team with just one woman, so the number of teams with at least two women is $$\binom{13}{10}-1-3\binom{10}9=\binom{13}{10}-31=255\;.$$ You need that factor of $3$ in the last term, because you can choose any one of the three women.
You can also count these teams directly. There are $\binom{10}7$ ways to choose a team with three women, and there are $\binom{10}8\binom32$ ways to choose a team with exactly two women, so the total is
$$\binom{10}7+\binom{10}8\binom32=120+45\cdot3=255\;.$$