Question

Our school's girls volleyball team has 14 players, including a set of 3 triplets: Missy, Lauren, and Liz. In how many ways can we choose 6 starters if the only restriction is that not all 3 triplets can be in the starting lineup?

Answer 1

A - not all 3 triplets can be in the starting lineup

A' - all 3 triplets can be in the starting lineup

$\displaystyle |\Omega|=\binom{14}{6}=\dfrac{14!}{6!8!}=\dfrac{9\cdot10\cdot11\cdot12\cdot13 \cdot14}{2\cdot3\cdot4\cdot5\cdot6}=3003\\ |A'|=\binom{11}{3}=\dfrac{11!}{3!8!}=\dfrac{9\cdot10\cdot11}{2\cdot3}=165\\ |A|=3003-165=2838\\\\ P(A)=\dfrac{2838}{3003}=\dfrac{86}{91}\approx95\%$