Question:
The number of ways of dividing 15 persons into 3 groups containing 3, 5 and 7 persons so that two particular persons are not included into the 5 persons group is
The number of ways of dividing 15 persons into 3 groups containing 3, 5 and 7 persons so that two particular persons are not included into the 5 persons group is
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When restrictions apply to group selections, carefully exclude restricted persons before selecting.
Updated On: Jun 6, 2025
- $\frac{117(11!)}{3!(7!)}$
- ${15 \choose 5} {10 \choose 3}$
- $90 \times \frac{13!}{7!}$
- ${15 \choose 5} {8 \choose 3}$
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The Correct Option is A
Solution and Explanation
Total persons = 15. Groups sizes = 3, 5, 7. Two particular persons cannot be in the 5-person group.
Step 1: Select 5-person group excluding these two persons: total eligible persons = 13 (15 - 2). Number of ways to select = ${13 \choose 5}$.
Step 2: From remaining 10 persons, select 3-person group: ${10 \choose 3}$.
Step 3: Remaining 7 persons form the last group automatically.
Step 4: Since groups are distinct by size, total ways = ${13 \choose 5} \times {10 \choose 3}$.
Calculate factorial form and simplify to get the answer $\frac{117(11!)}{3!(7!)}$.
Step 1: Select 5-person group excluding these two persons: total eligible persons = 13 (15 - 2). Number of ways to select = ${13 \choose 5}$.
Step 2: From remaining 10 persons, select 3-person group: ${10 \choose 3}$.
Step 3: Remaining 7 persons form the last group automatically.
Step 4: Since groups are distinct by size, total ways = ${13 \choose 5} \times {10 \choose 3}$.
Calculate factorial form and simplify to get the answer $\frac{117(11!)}{3!(7!)}$.
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