JEE Main 2025 Apr 7 Shift 1, Mathematics Q7: Combinations

JEE Main2025Apr 7, Shift 1Mathematics

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

A
$165$
B
$155$
C
$145$
D
$135$

Solution

The captain (1 batsman) and vice-captain (1 bowler) are already fixed. So from the remaining 6 batsmen and 5 bowlers we must choose 8 more players, keeping the totals at least 4 in each category.

Let the chosen team have $b$ batsmen and $w$ bowlers with $b + w = 10$ and $b, w \geq 4$ . So $(b, w) \in {(4, 6), (5, 5), (6, 4)}$ . But there are only 6 bowlers total, so $w \leq 6$ ; and only 7 batsmen, so $b \leq 7$ : all three cases are feasible.

After accounting for the fixed captain and vice-captain, we need to pick $(b - 1)$ additional batsmen from 6 and $(w - 1)$ additional bowlers from 5.

$(b, w)$	Extras from 6 batsmen	Extras from 5 bowlers	Count
$(4, 6)$	$(3 6)$	$(5 5)$	$20 \cdot 1 = 20$
$(5, 5)$	$(4 6)$	$(4 5)$	$15 \cdot 5 = 75$
$(6, 4)$	$(5 6)$	$(3 5)$	$6 \cdot 10 = 60$

Total $= 20 + 75 + 60 = 155$ .

$∴$ Option (2).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q7: Combinations

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