Maths Q8 – JEE-MAIN 2025

Home Maths Permutations and CombinationsQuestion #23

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

Clue 1 of 3: Identify Fixed Roles

Recognize that the captain (batsman) and vice-captain (bowler) are already designated and included in the team, reducing the number of players to be selected and the available pool.

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Step 1: Identify Fixed Players and Remaining Selection Task✦ Active

A team of 10 players is to be chosen. One batsman is designated as Captain, and one bowler as Vice-Captain. These two players are already included in the team. Therefore, we need to select $10 - 2 = 8$ more players.

The initial pool is 7 batsmen and 6 bowlers. After selecting 1 batsman (Captain) and 1 bowler (Vice-Captain), the remaining pool for selection is $7 - 1 = 6$ batsmen and $6 - 1 = 5$ bowlers.

Step 2: Determine Constraints for Remaining Players○ Expand

The final team must have at least 4 batsmen and at least 4 bowlers.

Since 1 batsman (Captain) is already in the team, we need to select at least $4 - 1 = 3$ more batsmen from the remaining 6 batsmen.

Since 1 bowler (Vice-Captain) is already in the team, we need to select at least $4 - 1 = 3$ more bowlers from the remaining 5 bowlers.

Let $b^{'}$ be the number of additional batsmen and $w^{'}$ be the number of additional bowlers. We must have $b^{'} + w^{'} = 8$ . The constraints are $3 \leq b^{'} \leq 6$ and $3 \leq w^{'} \leq 5$ .

Step 3: Calculate Combinations for Valid Cases○ Expand

We list the possible pairs of $(b^{'}, w^{'})$ that satisfy $b^{'} + w^{'} = 8$ and the given constraints:

Case 1: $b^{'} = 3$ , then $w^{'} = 5$ . (Valid: $3 \leq 3 \leq 6$ , $3 \leq 5 \leq 5$ )

Ways =^{6} C_{3} \times^{5} C_{5} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 1 = 20 \times 1 = 20

Case 2: $b^{'} = 4$ , then $w^{'} = 4$ . (Valid: $3 \leq 4 \leq 6$ , $3 \leq 4 \leq 5$ )

Ways =^{6} C_{4} \times^{5} C_{4} = \frac{6 \times 5}{2 \times 1} \times 5 = 15 \times 5 = 75

Case 3: $b^{'} = 5$ , then $w^{'} = 3$ . (Valid: $3 \leq 5 \leq 6$ , $3 \leq 3 \leq 5$ )

Ways =^{6} C_{5} \times^{5} C_{3} = 6 \times \frac{5 \times 4}{2 \times 1} = 6 \times 10 = 60

If $b^{'} = 6$ , then $w^{'} = 2$ , which violates $w^{'} \geq 3$ . So, no more cases.

Total number of ways to select the remaining 8 players is the sum of ways from these cases:

Total Ways = 20 + 75 + 60 = 155

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Maths Question 8 – JEE-MAIN 2025

Clue 1 of 3: Identify Fixed Roles

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