Question

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 _______.

Answer 1

Correct Answer: 665

Given conditions:

• \( s_1 \) cannot be in team X. 
• \( s_2 \) cannot be in team Y.

Case 1: \( s_2 \in X \)

The other member of team X is chosen from the remaining 7 students:

\[ \binom{7}{1} = 7 \]

The remaining 7 students are divided into teams Y and Z:

\[ \binom{7}{3} = 35 \]

The remaining 4 students are automatically in team Z (only 1 way to place them).

Total ways in this case:

\[ 7 \times 35 = 245 \]

Case 2: \( s_2 \notin X \)

The other 2 members of team X are chosen from the remaining 7 students (excluding \( s_2 \)):

\[ \binom{7}{2} = 21 \]

Now, \( s_2 \) is included in the remaining 7 students for teams Y and Z. Team Y (3 members) is chosen from the remaining 8 students, including \( s_2 \):

\[ \binom{8}{3} = 56 \]

The remaining 4 students are automatically in team Z (only 1 way to place them).

Total ways in this case:

\[ 21 \times 20 = 420 \]

Total Number of Ways

\[ 245 + 420 = 665 \]

Final Answer: 665

Answer 2

To solve the problem, count the number of ways to form three teams \(X\), \(Y\), and \(Z\) from 9 students \(s_1, s_2, ..., s_9\) with sizes 2, 3, and 4 respectively, under the restrictions:

• \(s_1\) cannot be in team \(X\)
• \(s_2\) cannot be in team \(Y\)

Step 1: Total students = 9

Step 2: Team sizes:
\[ |X| = 2, \quad |Y| = 3, \quad |Z| = 4 \]

Step 3: Strategy
Assign \(s_1\) and \(s_2\) first considering restrictions, then fill the remaining teams.

Step 4: Possible placements of \(s_1\) and \(s_2\)
- \(s_1\) can be in either \(Y\) or \(Z\) (not in \(X\)) → 2 choices
- \(s_2\) can be in either \(X\) or \(Z\) (not in \(Y\)) → 2 choices

Step 5: Consider cases based on placements of \(s_1\) and \(s_2\):

Case	\(s_1\)	\(s_2\)
1	\(Y\)	\(X\)
2	\(Y\)	\(Z\)
3	\(Z\)	\(X\)
4	\(Z\)	\(Z\)

Step 6: For each case, count the number of ways to complete the teams:
There are 7 remaining students after placing \(s_1\) and \(s_2\).

Recall team sizes:

| Team | Size | Members assigned so far (per case) | |-------|-------|-----------------------------------| | \(X\) | 2 | Depends on case | | \(Y\) | 3 | Depends on case | | \(Z\) | 4 | Depends on case |

Case 1: \(s_1 \in Y\), \(s_2 \in X\)
- \(X\): 1 member assigned (\(s_2\)), need 1 more from remaining 7
- \(Y\): 1 member assigned (\(s_1\)), need 2 more from remaining 7
- \(Z\): 0 assigned, need 4 from remaining 7

Number of ways:

\[ \binom{7}{1} \times \binom{6}{2} \times \binom{4}{4} = 7 \times 15 \times 1 = 105 \]

Case 2: \(s_1 \in Y\), \(s_2 \in Z\)
- \(X\): 0 assigned, need 2 from 7
- \(Y\): 1 assigned (\(s_1\)), need 2 from 7
- \(Z\): 1 assigned (\(s_2\)), need 3 from 7

Number of ways:

\[ \binom{7}{2} \times \binom{5}{2} \times \binom{3}{3} = 21 \times 10 \times 1 = 210 \]

Case 3: \(s_1 \in Z\), \(s_2 \in X\)
- \(X\): 1 assigned (\(s_2\)), need 1 from 7
- \(Y\): 0 assigned, need 3 from 7
- \(Z\): 1 assigned (\(s_1\)), need 3 from 7

Number of ways:

\[ \binom{7}{1} \times \binom{6}{3} \times \binom{3}{3} = 7 \times 20 \times 1 = 140 \]

Case 4: \(s_1 \in Z\), \(s_2 \in Z\)
- \(X\): 0 assigned, need 2 from 7
- \(Y\): 0 assigned, need 3 from 7
- \(Z\): 2 assigned (\(s_1, s_2\)), need 2 from 7

Number of ways:

\[ \binom{7}{2} \times \binom{5}{3} \times \binom{2}{2} = 21 \times 10 \times 1 = 210 \]

Step 7: Total number of ways:
\[ 105 + 210 + 140 + 210 = 665 \]

Final Answer:
\[ \boxed{665} \]