The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

Solution

Maths sir · 4 steps

First choose which 5 boxes get letters, then arrange the 5 letters in them in 5! ways.

1. Read the figure. The top row has 3 boxes, the middle row has 3 boxes (the wide one counts as one box) and the bottom row has 2 boxes. That makes 8 boxes in all.
2. Count all choices. Each box holds at most one letter, so the 5 letters fill 5 different boxes. Ignoring the row condition, there are C(8, 5) = 56 ways to choose the boxes.

   Row left empty	Boxes still available	Ways to choose 5 boxes
   Top row (3 boxes)	3 + 2 = 5	C(5, 5) = 1
   Middle row (3 boxes)	3 + 2 = 5	C(5, 5) = 1
   Bottom row (2 boxes)	3 + 3 = 6	C(6, 5) = 6

3. Remove bad choices. Two rows cannot both be empty, because the other row has at most 3 boxes, too few for 5 letters. So exactly 1 + 1 + 6 = 8 choices leave a row empty, and 56 - 8 = 48 choices are allowed.

   48 × 5! = 48 × 120 = 5760

4. So. The 5 distinct letters can be arranged in the chosen boxes in 5! = 120 ways each, giving 5760 ways in total, which is option (d).

The idea to keep. For distinct objects in distinct boxes, split the job: choose the boxes first, then multiply by the arrangements. For a condition like no row empty, subtract the selections that break it.

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