In a group of 3 girls and 4 boys, there are two boys B₁ and B₂. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B₁ and B₂ are not adjacent to each other, is :

Solution

Maths sir · 4 steps

Tie each group into a block, arrange inside each block, and subtract the cases where B1 and B2 touch.

1. Arrange the blocks. Treat the 3 girls as one block and the 4 boys as one block. The two blocks can stand in 2! = 2 orders: girls first or boys first.
2. Girls inside. The 3 girls can be ordered among themselves in 3! = 6 ways.
3. Boys inside. All orders of the 4 boys number 4! = 24. To count orders with B₁ and B₂ adjacent, glue them into one unit: 3! × 2! = 12. That leaves 24 - 12 = 12 orders where they are apart.

   2! × 3! × (4! - 3! × 2!) = 2 × 6 × 12 = 144

4. So. The number of valid queues is 144, which is option (d).

The idea to keep. For 'together' conditions, glue the group into one unit and multiply by its internal arrangements. For 'not together', subtract the glued count from the total.

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