Question

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

Answer 1

Answer:

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

Step-by-step explanation:

The arrangements can be determined in three steps:

• determine the man/woman sequence
• assign the 10 to the "man" positions
• assign the 6 women to the "woman" positions

For the man/woman sequence, put down 10 "M" labels first.  Now we need to put 6 "W" labels amongst these so that no two of the "W" labels are together.  There are 11 gaps between "M" labels (including before and after them) and we just need to choose 6 of these to put one "W" label in each.  The number of ways of doing this is $\displaystyle\binom{11}{6}=\frac{11!}{6!\,5!}$

The number of ways of assigning the 10 men to the 10 "M" labelled positions is just the number of ways of permuting the 10 men.... 10!

Similarly, the number of ways of assigning women to the "W" labelled positions is 6!.

So the total number of ways of seating is

$\displaystyle\frac{11!}{6!\,5!}\times10!\times6! = \frac{11!\,10!}{5!}$