From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?

We first count the number of committee in which
(i). Mr. Y is a member
(ii). the ones in which he is not
Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).
We can choose 1 more in5+ $2C1$=7 ways.
Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in $9C3$=84 ways.
Thus, total number of ways is 7+84= 91 ways.