In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
A. $$\frac{{6!}}{{3! \times 3! \times 3!}}$$
B. $$\frac{{6!}}{{2! \times 2!}}$$
C. $$\frac{{6!}}{{2! \times 2! \times 2!}}$$
D. $$\frac{{6!}}{{2! \times 3!}}$$
Answer: Option C
Solution(By Examveda Team)
6 people can be made to stand in a line in 6! Ways. However, the problem introduces a constraint that no man stands in a position that is ahead of his wife. For any 2 given positions out of the 6 occupied by a man and his wife, the pair cannot rearrange amongst themselves in 2! Ways as the wife has to be in a position ahead of the man. Only one of the 2! arrangements is allowed. As there are 3 couples in the group, the total number of ways gets reduced by a factor of (2! × 2! × 2!) Hence, the total number of ways, = $$\frac{{6!}}{{2! \times 2! \times 2!}}$$Related Questions on Permutation and Combination
A. 3! 4! 8! 4!
B. 3! 8!
C. 4! 4!
D. 8! 4! 4!
A. 7560,60,1680
B. 7890,120,650
C. 7650,200,4444
D. None of these
A. 8 × 9!
B. 8 × 8!
C. 7 × 9!
D. 9 × 8!
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