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 _______.
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 _______.
Updated On: Mar 8, 2025
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Correct Answer: 665
Solution and Explanation
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
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