JEE Main 2025 Jan 24 Shift 2, Mathematics Q16: Combinations

JEE Main2025Jan 24, Shift 2Mathematics

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

A
$8575$
B
$9100$
C
$8925$
D
$8750$

Solution

Constraints: total invited $=$ 4 boys $+$ 4 girls $=$ 8 people; from Group A $=$ 5, from Group B $=$ 3.

Let $g_{A}$ be the number of girls chosen from Group A. Then boys from $A = 5 - g_{A}$ , girls from $B = 4 - g_{A}$ , boys from $B = 3 - (4 - g_{A}) = g_{A} - 1$ .

Constraints on counts:

Group A girls: $0 \leq g_{A} \leq 3$
Group A boys: $0 \leq 5 - g_{A} \leq 7 \Rightarrow g_{A} \geq 0$ (and $g_{A} \leq 5$ )
Group B girls: $0 \leq 4 - g_{A} \leq 5 \Rightarrow g_{A} \geq 0$ (and $g_{A} \leq 4$ )
Group B boys: $0 \leq g_{A} - 1 \leq 6 \Rightarrow g_{A} \geq 1$

So $g_{A} \in {1, 2, 3}$ . Compute each case:

$g_{A}$	Composition	Count
1	A: 4B, 1G; B: 0B, 3G	$(4 7) (1 3) (0 6) (3 5) = 35 \cdot 3 \cdot 1 \cdot 10 = 1050$
2	A: 3B, 2G; B: 1B, 2G	$(3 7) (2 3) (1 6) (2 5) = 35 \cdot 3 \cdot 6 \cdot 10 = 6300$
3	A: 2B, 3G; B: 2B, 1G	$(2 7) (3 3) (2 6) (1 5) = 21 \cdot 1 \cdot 15 \cdot 5 = 1575$

Total $= 1050 + 6300 + 1575 = 8925$ .

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q16: Combinations

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