Combinations
The number of ways to select a cricket team of 11 players from 15 players, where 2 particular players must be included, is
Select the correct option:
Solution
715
- Analyze Constraints: We need a team of 11. 2 players are already selected (mandatory inclusion).
- Selection Task: We need to pick the remaining 11−2=9 players.
- Available Pool: We pick these from the remaining 15−2=13 players.
- Apply Combination: (913).
- Symmetry Property: (913)=(413).
- Calculation:
- (413)=4×3×2×113×12×11×10=213×11×10=13×11×5=715.
- Result: 715 ways.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More combinations Practice Questions
Using the standard combinatorial identity, the value of the sum C(8,0) + C(8,1) + C(8,2) + ... + C(8...
Using the standard combinatorial identity, the value of the sum C(8,0) + C(8,1) + C(8,2) + ... + C(8...
If the combinatorial equation C(n, 2) equals 28 holds for a positive integer n, then the value of n ...
If the combinatorial equation C(n, 2) equals 28 holds for a positive integer n, then the value of n ...
From a group of 7 men and 6 women, the number of ways to form a committee consisting of exactly 3 me...
From a group of 7 men and 6 women, the number of ways to form a committee consisting of exactly 3 me...
The number of ways to distribute 10 identical chocolates among 3 distinct children such that each ch...
The number of ways to distribute 10 identical chocolates among 3 distinct children such that each ch...
If C(n, 4) = C(n, 6), then n equals
If C(n, 4) = C(n, 6), then n equals
About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- combinations
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
715
- Analyze Constraints: We need a team of 11. 2 players are already selected (mandatory inclusion).
- Selection Task: We need to pick the remaining 11−2=9 players.
- Available Pool: We pick these from the remaining 15−2=13 players.
- Apply Combination: (913).
- Symmetry Property: (913)=(413).
- Calculation:
- (413)=4×3×2×113×12×11×10=213×11×10=13×11×5=715.
- Result: 715 ways.
This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of combinations. It appeared in the 2025 exam.
Looking for more practice? Explore all mathematics questions or browse permutations and combinations questions on RankGuru.