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, = $$\frac{{6!}}{{2! \times 2! \times 2!}}$$This Question Belongs to Arithmetic Ability >> Permutation And Combination
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