JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?
- (a) 2! 3! 4!
- (b) (3!)³.(4!)
- (c) 3! (4!)³
- (d) (3!)².(4!)
Answer and solution
Answer (b) (3!)³.(4!)
Solution
Maths sir · 3 steps
Glue each family into one block, arrange the blocks, then arrange people inside every block.
- Make blocks. Each family must sit together, so treat each family as one block. That gives 3 blocks in a row, which can be ordered in 3! ways.
- Arrange inside. Inside its block, each three-member family can sit in 3! ways and the four-member family in 4! ways. These choices are independent of each other and of the block order.
3!_(blocks) × 3! × 3! × 4! = (3!)³ × 4!
- So. The number of seatings is (3!)³ × 4! = 216 × 24 = 5184, which is option (b).
The idea to keep. For 'stay together' problems, multiply two things: the ways to order the blocks, and the ways to arrange people inside each block. Never drop either factor.
Answer from the NTA final answer key. Solution by PrepWiser.