The number of ways of choosing a committee of4 women

 and 5 men from 10 women and 9 men, if Mr. A refuses to

 serve on the committee when Ms. B is a member of the

 committee, is 

The number of ways of choosing a committee if 

there is no restriction is 

  ${ }^{10}{C}_4{\cdot }^9{C}_5=\frac{10!}{4!6!}\cdot \frac{9!}{4!5!}=26460$ 

The number of ways of choosing the committee if   both Mr. A 

and M . Bare included in the committee is  ${ }^9{C}_3{\cdot }^8{C}_4=5880$

Therefore, the number of ways of choosing the committee 

when Mr. $A$  and Ms. $B$arc not together = 26480- 5880 = 20580.