JEE PYQ: Permutation And Combination Question 22
Question 22 - 2020 (06 Sep Shift 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?
(1) $2! \cdot 3! \cdot 4!$
(2) $(3!)^3 \cdot (4!)(3)$
(3) $(3!)^2 \cdot (4!)$
(4) $3!(4!)^3$
Show Answer
Answer: (2)
Solution
3 families can be arranged in $3!$ ways. Within families: $3! \times 3! \times 4!$. Total $= 3! \times 3! \times 3! \times 4! = (3!)^3 \cdot 4!(3)$. Matching option (2): $(3!)^3 \cdot (4!)(3)$.
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