Question 11

Total number of ways to divide 9 students into teams of 2, 3, and 4 is $N = \frac{9!}{2!3!4!} = 1260$. Let $A$ be the set of ways where $s_1$ is in team $X$. $n(A) = \binom{1}{1} \times \binom{8}{1} \times \binom{7}{3} \times \binom{4}{4} = 8 \times 35 = 280$. Let $B$ be the set of ways where $s_2$ is in team $Y$. $n(B) = \binom{8}{2} \times \binom{1}{1} \times \binom{6}{2} \times \binom{4}{4} = 28 \times 15 = 420$. We want to find $N - n(A \cup B) = N - [n(A) + n(B) - n(A \cap B)]$. $A \cap B$ is the set of ways where $s_1 \in X$ and $s_2 \in Y$. Remaining slots: Team $X$ needs 1 more, Team $Y$ needs 2 more, Team $Z$ needs 4. Remaining students: 7. $n(A \cap B) = \binom{7}{1} \times \binom{6}{2} \times \binom{4}{4} = 7 \times 15 = 105$. Ways = $1260 - [280 + 420 - 105] = 1260 - 595 = 665$.