Thank you for reporting, we will resolve it shortly
Q.
Number of ways in which $3$ men and their wives can be arranged in a line such that none of the $3$ men stand in a position that is ahead of his wife, is
Permutations and Combinations
Solution:
Treat $H _{1} W _{1}$ as alike objects say $XX$ $H _{2} W _{2}$ as $YY$ and $H _{3} W _{3}$ as $ZZ$
Now arrangements of $X X Y Y Z Z$ i.e. $\frac{6 !}{2 ! 2 ! 2 !}$ is this required number of ways (think!)
Alter $:{ }^{6} C _{2}\left(\right.$ for $\left.H _{1} W _{1}\right)$ and ${ }^{4} C _{2}\left(\right.$ for $\left.H _{2} W _{2}\right) \times \times \times \times \times \times$
Hence total ways $\left({ }^{6} C _{2} \cdot 1\right)\left({ }^{4} C _{2} \cdot 1\right)=\frac{6 !}{2 ! 2 ! 2 !}$
