Number of Ways to Divide Students into Teams with Specific Restrictions

To find the number of ways to divide the $9$ students $s_1, s_2, \dots, s_9$ into three teams $X, Y$, and $Z$ of sizes $2, 3$, and $4$ respectively, subject to the conditions that $s_1 \notin X$ and $s_2 \notin Y$, we can use two different approaches: the Principle of Inclusion-Exclusion (PIE) and direct casework.

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Method 1: Principle of Inclusion-Exclusion (PIE)

Step 1: Total number of unrestricted ways The total number of ways to divide $9$ students into teams $X, Y, Z$ of sizes $2, 3, 4$ without any restrictions is: $N_{\text{total}} = \binom{9}{2} \times \binom{7}{3} \times \binom{4}{4}$ $N_{\text{total}} = 36 \times 35 \times 1 = 1260$

Step 2: Number of ways where restriction is violated Let $A$ be the set of distributions where student $s_1$ is selected in team $X$. Let $B$ be the set of distributions where student $s_2$ is selected in team $Y$.

• Finding $N(A)$ (where $s_1 \in X$): Since $s_1$ is fixed in team $X$, we need to choose $1$ more student for $X$ from the remaining $8$ students, then $3$ students for $Y$ from the remaining $7$ students, and the remaining $4$ students go to $Z$: $N(A) = \binom{8}{1} \times \binom{7}{3} \times \binom{4}{4} = 8 \times 35 \times 1 = 280$

• Finding $N(B)$ (where $s_2 \in Y$): Since $s_2$ is fixed in team $Y$, we need to choose $2$ more students for $Y$ from the remaining $8$ students, then $2$ students for $X$ from the remaining $6$ students, and the remaining $4$ students go to $Z$: $N(B) = \binom{8}{2} \times \binom{6}{2} \times \binom{4}{4} = 28 \times 15 \times 1 = 420$

• Finding $N(A \cap B)$ (where $s_1 \in X$ and $s_2 \in Y$): With $s_1 \in X$ and $s_2 \in Y$, we select $1$ more student for $X$ from the remaining $7$ students, $2$ more students for $Y$ from the remaining $6$ students, and the rest $4$ go to $Z$: $N(A \cap B) = \binom{7}{1} \times \binom{6}{2} \times \binom{4}{4} = 7 \times 15 \times 1 = 105$

Step 3: Applying PIE The number of valid team formations is given by $N(\bar{A} \cap \bar{B})$: $N(\bar{A} \cap \bar{B}) = N_{\text{total}} - N(A) - N(B) + N(A \cap B)$ $N(\bar{A} \cap \bar{B}) = 1260 - 280 - 420 + 105 = 665$

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Method 2: Casework Analysis

Since $s_1 \notin X$ and $s_2 \notin Y$, $s_1$ can belong to $Y$ or $Z$, and $s_2$ can belong to $X$ or $Z$. This gives four mutually exclusive cases for the pair of team assignments $(s_1, s_2)$:

1. Case 1: $s_1 \in Y$ and $s_2 \in X$

   • Select $1$ remaining student for $X$ from $7$: $\binom{7}{1} = 7$
   • Select $2$ remaining students for $Y$ from $6$: $\binom{6}{2} = 15$
   • Select $4$ remaining students for $Z$ from $4$: $\binom{4}{4} = 1$ $\text{Ways}_1 = 7 \times 15 \times 1 = 105$

2. Case 2: $s_1 \in Y$ and $s_2 \in Z$

   • Select $2$ students for $X$ from $7$: $\binom{7}{2} = 21$
   • Select $2$ remaining students for $Y$ from $5$: $\binom{5}{2} = 10$
   • Select $3$ remaining students for $Z$ from $3$: $\binom{3}{3} = 1$ $\text{Ways}_2 = 21 \times 10 \times 1 = 210$

3. Case 3: $s_1 \in Z$ and $s_2 \in X$

   • Select $1$ remaining student for $X$ from $7$: $\binom{7}{1} = 7$
   • Select $3$ remaining students for $Y$ from $6$: $\binom{6}{3} = 20$
   • Select $3$ remaining students for $Z$ from $3$: $\binom{3}{3} = 1$ $\text{Ways}_3 = 7 \times 20 \times 1 = 140$

4. Case 4: Both $s_1 \in Z$ and $s_2 \in Z$

   • Select $2$ students for $X$ from $7$: $\binom{7}{2} = 21$
   • Select $3$ remaining students for $Y$ from $5$: $\binom{5}{3} = 10$
   • Select $2$ remaining students for $Z$ from $2$: $\binom{2}{2} = 1$ $\text{Ways}_4 = 21 \times 10 \times 1 = 210$

Total Ways: $\text{Total} = 105 + 210 + 140 + 210 = 665$

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The number of ways to form such teams is 665.