Question

The number of ways in which a cricket team of 11 members can be formed out of 6 batsmen, 6 bowlers, 4 all-rounders, and 4 wicket-keepers by selecting at least 4 batsmen, at least 3 bowlers, at least 2 all-rounders, and only one wicket-keeper is

• A. 11560
• B. 6480
• C. 7680
• D. 13080

Hint:

List all valid combinations summing to the total (e.g., 11 players) while satisfying minimum constraints, and compute each using combinations: $\binom{n}{r}$. Verify the sum matches an option.

Answer 1

The Correct Option is D

We need to select 11 players with: - At least 4 batsmen (from 6) - At least 3 bowlers (from 6) - At least 2 all-rounders (from 4) - Exactly 1 wicket-keeper (from 4) The total number of players must be 11. Possible combinations satisfying the constraints: 1. 4 batsmen, 4 bowlers, 2 all-rounders, 1 wicket-keeper: $4 + 4 + 2 + 1 = 11$ \[ \binom{6}{4} \cdot \binom{6}{4} \cdot \binom{4}{2} \cdot \binom{4}{1} = 15 \cdot 15 \cdot 6 \cdot 4 = 5400 \] 2. 5 batsmen, 3 bowlers, 2 all-rounders, 1 wicket-keeper: $5 + 3 + 2 + 1 = 11$ \[ \binom{6}{5} \cdot \binom{6}{3} \cdot \binom{4}{2} \cdot \binom{4}{1} = 6 \cdot 20 \cdot 6 \cdot 4 = 2880 \] 3. 4 batsmen, 3 bowlers, 3 all-rounders, 1 wicket-keeper: $4 + 3 + 3 + 1 = 11$ \[ \binom{6}{4} \cdot \binom{6}{3} \cdot \binom{4}{3} \cdot \binom{4}{1} = 15 \cdot 20 \cdot 4 \cdot 4 = 4800 \] Total ways: \[ 5400 + 2880 + 4800 = 13080 \] The original solution incorrectly computed the third case as $\binom{4}{3}$ instead of $\binom{4}{2}$. Option (4) is correct. Options (1), (2), and (3) do not match the total.