A group of 9 students$, s_{1}, s_{2}, \dots , s_{9}$ is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that $s_{1}$ cannot be selected for the team X, and $s_{2}$ cannot be selected for the team Y. Then the number of ways to form such teams, is ______.

A group of 9 students, s₁, s₂, ..... s₉ is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that s₁ cannot be selected for the team X, and s₂ cannot be selected for the team Y. Then the number of ways to form such teams, is ______.

$\text{Given: } 9 \text{ students } s_1,\dots,s_9 \to X,Y,Z \text{ of sizes } 2,3,4$

$s_1 \notin X,\qquad s_2 \notin Y$

$\text{To find: the number of ways to form the teams}$

$1.\ \text{count all ways to fill } X(2),Y(3),Z(4)$

$2.\ \text{subtract cases with } s_1\in X \text{ or } s_2\in Y$

$3.\ \text{inclusion--exclusion, then verify by cases}$

$\text{choose } X: \binom{9}{2},\quad \text{then } Y: \binom{7}{3}$

$Z \text{ gets the remaining } 4 \text{ automatically}$

$\text{Total} = \binom{9}{2}\binom{7}{3} = 36\cdot35 = 1260$

$A = \{s_1 \in X\},\qquad B = \{s_2 \in Y\}$

$|A| = \binom{8}{1}\binom{7}{3} = 8\cdot35 = 280$

$|B| = \binom{8}{2}\binom{6}{2} = 28\cdot15 = 420$

$|A\cap B| = \binom{7}{1}\binom{6}{2} = 7\cdot15 = 105$

$|A\cup B| = 280+420-105 = 595$

$\text{Valid} = 1260 - 595 = 665$

$\text{check by cases on } (s_1,s_2)$

$(Y,X): 7\cdot15 = 105,\qquad (Y,Z): 21\cdot10 = 210$

$(Z,X): 7\cdot20 = 140,\qquad (Z,Z): 21\cdot10 = 210$

$105+210+140+210 = 665$

$\boxed{665}$