There are ten boys B₁, B₂, ......., B₁₀ and five girls G₁, G₂, ........, G₅ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B₁ and B₂ together should not be the members of a group, is ___________.

Solution

Maths sir · 3 steps

Count every group first, then throw out the ones where B1 and B2 sit together.

1. All groups. Choose 3 of the 10 boys and 3 of the 5 girls. That gives C(10, 3) × C(5, 3) = 120 × 10 = 1200 groups with no restriction.
2. Bad groups. If B₁ and B₂ are both in, only one more boy is needed, from the other 8 boys. The girls are still any 3 of the 5, so there are C(8, 1) × C(5, 3) = 8 × 10 = 80 such groups.

   C(10, 3)C(5, 3) - C(8, 1)C(5, 3) = 1200 - 80

3. So. The number of groups without both B₁ and B₂ is 1200 - 80 = 1120.

The common slip: 1192. Subtracting only C(8, 1) = 8 and forgetting that the 3 girls still have to be chosen in C(5, 3) = 10 ways gives 1192, not 1120.

The idea to keep. When a condition says two items must not both be chosen, count all selections and subtract the ones that contain both. Keep every other part of the choice unchanged in the subtracted count.

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