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 1 s 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 _______.

The correct answer is 665.

To find the number of ways to form the teams under the given constraints, we can use the Principle of Inclusion-Exclusion.

Step 1: Calculate the total number of ways to divide 9 students into three teams without any restrictions.
We need to form three teams:

• Team X of size 2
• Team Y of size 3
• Team Z of size 4

$\mathrm{Total}=\frac{9!}{2!\times 3!\times 4!}=\frac{9\times 8\times 7\times 6\times 5}{2\times 1\times 6}=1260$

Step 2: Calculate the number of ways that violate the constraints.
Let A be the event that student S₁ is selected for team X.
Let B be the event that student S₂ is selected for team Y.

Case 2a: S₁ is in team X (Event A)
Since S₁ is already placed in X, we need to choose 1 more student for team X from the remaining 8 students, then 3 students for team Y from the remaining 7, and the remaining 4 students will go to team Z.

$n\left(A\right)=\left(\genfrac{}{}{0ex}{}{8}{1}\right)\times \left(\genfrac{}{}{0ex}{}{7}{3}\right)\times \left(\genfrac{}{}{0ex}{}{4}{4}\right)=8\times 35\times 1=280$

Case 2b: S₂ is in team Y (Event B)
Since S₂ is already placed in Y, we need to choose 2 students for team X from the remaining 8 students, then 2 more students for team Y from the remaining 6, and the remaining 4 students will go to team Z.

$n\left(B\right)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)\times \left(\genfrac{}{}{0ex}{}{6}{2}\right)\times \left(\genfrac{}{}{0ex}{}{4}{4}\right)=28\times 15\times 1=420$

Case 2c: Both S₁ is in team X and S₂ is in team Y (Event A ∩ B)
Since S₁ is placed in X and S₂ is placed in Y, we need to choose 1 more student for team X from the remaining 7 students, then 2 more students for team Y from the remaining 6, and the remaining 4 students will go to team Z.

$n(A\cap B)=\left(\genfrac{}{}{0ex}{}{7}{1}\right)\times \left(\genfrac{}{}{0ex}{}{6}{2}\right)\times \left(\genfrac{}{}{0ex}{}{4}{4}\right)=7\times 15\times 1=105$

Step 3: Apply the Principle of Inclusion-Exclusion.
The number of invalid ways where either S₁ is in team X or S₂ is in team Y is:

$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$

$n(A\cup B)=280+420-105=595$

Step 4: Subtract the invalid ways from the total ways.
The number of valid ways to form the teams is:

$\mathrm{Number\; of\; ways}=\mathrm{Total}-n(A\cup B)$

$\mathrm{Number\; of\; ways}=1260-595=665$