Aptitude Discussion

Q. |
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. |
$\dfrac{6!}{3! × 3! × 3!}$ |

✖ B. |
$\dfrac{6!}{2! × 2!}$ |

✔ C. |
$\dfrac{6!}{2! × 2! × 2!}$ |

✖ D. |
$\dfrac{6!}{2! × 3!}$ |

**Solution:**

Option(**C**) is correct

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 re-arrange 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,

$ = \dfrac{6!}{2! × 2! × 2!}$

**Shubham Kumar**

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Does anybody have an easier explanation to the question apart from the given one?