10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is :

Question

$$\begin {array} {1 1} (a)\;11!\;10! & \quad (b)\;\frac{11 !}{6!\;5!} \\ (c)\;\frac{10 !\; 9!}{5!} & \quad (d)\;\frac{11 !\; 10 !}{5 !} \end {array}$$