Question

10 men and 6 women are to be seated in a row so that no 2 women sit together. The number of ways they can be seated is

• A. $11!10!$
• B. $\frac{11!}{6!5!}$
• C. $\frac{10!9!}{5!}$
• D. $\frac{11!10!}{5!}$

Answer 1

Answer: (D)

$\frac{11!10!}{5!}$

$10men+6women$
M M M .....
these are 11 places where women can be arranged .

$10!\times 11C_6\times 6!=\frac{10!\times 11!}{6!\times 5!}\times 6!$

$=\frac{10!\times 11!}{5!}$