JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Permutations and Combinations

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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) $(3!)^{3} \cdot (4!)$ ✓ Correct
(C) $(3!)^{2} \cdot (4!)$
(D) $3! (4!)^{3}$

Answer: (B)

Solution

Keep each family's members together by treating every family as a single block, so three blocks are arranged in a row in

3!

ways. Within the blocks the members can be ordered among themselves:

3!

ways in each of the two three-member families and

4!

ways in the four-member family. Multiplying, the total number of seatings is

3! \times 3! \times 3! \times 4! = (3!)^{3} (4!)

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option

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