Permutations and Combinations — JEE Main · 2025 · 24 Jan 2025 · Shift 2 Question with Solution

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✓ Correct
(D)8750

Answer: (C)

Solution

Group A has 7 boys and 3 girls; Group B has 6 boys and 5 girls. We need 4 boys and 4 girls overall, with exactly 5 people from Group A and 3 from Group B.

Let

g

be the number of girls picked from Group A, so

5 - g

boys come from Group A. Since Group A has only 3 girls,

g \leq 3

. The remaining girls needed,

4 - g

, and remaining boys,

4 - (5 - g) = g - 1

, must come from Group B, and

g - 1 \geq 0

forces

g \geq 1

. So

g \in {1, 2, 3}

g = 1

: from A, 1 girl and 4 boys; from B, 3 girls and 0 boys.

(1 3) (4 7) (3 5) (0 6) = 3 \times 35 \times 10 \times 1 = 1050.

g = 2

: from A, 2 girls and 3 boys; from B, 2 girls and 1 boy.

(2 3) (3 7) (2 5) (1 6) = 3 \times 35 \times 10 \times 6 = 6300.

g = 3

: from A, 3 girls and 2 boys; from B, 1 girl and 2 boys.

(3 3) (2 7) (1 5) (2 6) = 1 \times 21 \times 5 \times 15 = 1575.

Adding the three mutually exclusive cases:

1050 + 6300 + 1575 = 8925.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Permutations and Combinations

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