There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is ________.

Solution

Maths sir · 3 steps

Fix how many men come from Group A, and the counts for Group B are forced. Then add up the cases.

1. Set up cases. Say r men and 4-r women come from Group A. To get 4 men in total, Group B must give 4-r men, so it gives r women. Here r runs from 0 to 4.
2. Count one case. Group A gives C(4, r)C(5, 4-r) ways. Group B gives C(5, 4-r)C(4, r) ways, which is the same number. Multiply the two groups, because both selections happen together.

   N = Σᵣ₌₀⁴ [C(4, r)C(5, 4-r)]²

   Men from A (r)	Ways in A	Ways in B	Product
   0	1 × 5 = 5	5 × 1 = 5	25
   1	4 × 10 = 40	10 × 4 = 40	1600
   2	6 × 10 = 60	10 × 6 = 60	3600
   3	4 × 5 = 20	5 × 4 = 20	400
   4	1 × 1 = 1	1 × 1 = 1	1

3. So. Add the cases: 25 + 1600 + 3600 + 400 + 1 = 5626.

The common slip: 126. Counting only Group A's choices and forgetting Group B gives 5 + 40 + 60 + 20 + 1 = 126, not 5626. Each case must be multiplied by Group B's ways too.

The idea to keep. When two selections must meet a combined total, fix the split in one group. The other group's split is then forced. Multiply within a case, add across cases.

Answer from the NTA final answer key. Solution by PrepWiser.