We need to divide 9 distinct students into three labeled teams:

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

with restrictions:

• $s_1 \notin X$
• $s_2 \notin Y$

We count the number of valid divisions.

1. Total number of ways without restrictions

Since the teams $X,Y,Z$ are distinct (labeled), the number of ways is

2. Subtract arrangements violating the restrictions

Let

• $A$ = arrangements where $s_1 \in X$
• $B$ = arrangements where $s_2 \in Y$

We want

3. Count $|A|$: $s_1 \in X$

Team $X$ has size $2$, and one member is already $s_1$. So choose the other member of $X$ from the remaining 8 students:

Now 7 students remain, from which team $Y$ of size 3 can be chosen in

ways. Then team $Z$ is fixed.

Hence,

4. Count $|B|$: $s_2 \in Y$

Team $Y$ has size $3$, and one member is already $s_2$. Choose the remaining 2 members of $Y$ from the other 8 students:

Now 6 students remain, from which team $X$ of size 2 can be chosen in

ways. Then team $Z$ is fixed.

Thus,

5. Count $|A\cap B|$: both violations occur

Here,

• $s_1 \in X$
• $s_2 \in Y$

Choose the second member of $X$ from the remaining 7 students (everyone except $s_1,s_2$ and possibly including no one else):

Now choose the remaining 2 members of $Y$ from the remaining 6 students:

Then team $Z$ is fixed.

So,

6. Apply inclusion-exclusion

Therefore,

Compute:

Hence the required number of ways is

7. Comparison with stored answer

Stored correct answer = $665$.

Our derived answer matches the stored answer.