Group A: 7 Boys 3 Girls, Group B: 6 Boys 5 Girls — Picnic Selection

JEE Main 2025 — 24 January, Evening Shift. Previous Year Question.

Problem

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:

Key insight. The constraint “5 from group A, 3 from group B” doesn’t say how those 5 (or 3) split between boys and girls — only the final totals (4 boys, 4 girls overall) are fixed. That leaves multiple valid ways to split boys and girls within each group’s quota, and each split has to be counted as a separate case.

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Approach

Since exactly 5 people must come from group A and exactly 3 from group B, the question becomes: in how many ways can that 5-and-3 split still add up to exactly 4 boys and 4 girls overall? Trying every possible number of boys taken from group A (from 4 down to 2, since taking fewer would require more girls than group A has) generates a small, manageable set of cases, each counted with combinations and then summed.

Solution

Step 1 — Set up the possible cases

Group A must contribute 5 people; group B must contribute 3. If group A contributes $b$ boys, it contributes $5-b$ girls — but group A only has 3 girls, so $5-b\le 3$, meaning $b\ge 2$. Since group A only has 7 boys (more than enough) and the total boys needed is 4, $b$ can be at most 4. So $b \in \{2,3,4\}$, giving three cases:

• Case 1: 4 boys + 1 girl from A, 0 boys + 3 girls from B
• Case 2: 3 boys + 2 girls from A, 1 boy + 2 girls from B
• Case 3: 2 boys + 3 girls from A, 2 boys + 1 girl from B

(Each case is forced to add up to 4 boys and 4 girls overall, since group B’s boy/girl split is whatever makes the totals work.)

Step 2 — Count Case 1

$${}^7C_4 \times {}^3C_1 \times {}^6C_0 \times {}^5C_3 = 35\times 3\times 1\times 10 = 1050$$

Step 3 — Count Case 2

$${}^7C_3 \times {}^3C_2 \times {}^6C_1 \times {}^5C_2 = 35\times 3\times 6\times 10 = 6300$$

Step 4 — Count Case 3

$${}^7C_2 \times {}^3C_3 \times {}^6C_2 \times {}^5C_1 = 21\times 1\times 15\times 5 = 1575$$

Step 5 — Add all cases

$$1050 + 6300 + 1575 = 8925$$

Answer

Common mistakes

• Assuming the 5-from-A and 3-from-B split must divide boys and girls evenly and only checking one case — the problem allows several different internal splits, and missing cases undercounts the total significantly.
• Forgetting the boundary case where group A contributes 0 boys (all 5 as girls) is impossible, since group A only has 3 girls — not checking this leads to including an invalid case with a negative or unavailable count.

Practise next

• If Group A has 5 boys and 4 girls, and Group B has 4 boys and 6 girls, find the number of ways to invite 3 boys and 3 girls with exactly 4 from Group A and 2 from Group B, using the same case-by-case method.