# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. 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?
 ✖ A. 138 ✖ B. 128 ✖ C. 112 ✔ D. 91

Solution:
Option(D) is correct

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 in $^{5+2}C_1$=7$ways. case (ii): If Mr. Y is not a member then we left with$(6+4-1)$people. we can select 3 from 9 in${^9C_3} = 84$ways. Thus, total number of ways is$7+84 =\textbf{91}$ways. ## (3) Comment(s) Shishir () There will be two cases 1: Mrs Z is not a member in that case Mr Y will not be willing to join the committee. So we are left with 8 person. So no of ways in which 3 persons can be selected= 8C3=56 2:Mrs Z is a member Now under this case we have total 9 person and 2 slots. So number of ways of choosing 2 persons= 9C2 but this will include the case in which Mr Y and Mrs X are choosen. So to satisfy the imposed condition No of ways is =9C2-1=35 Therefore total no of ways =35 + 56=91 Priyanka () in case (ii) if Mr.Y is not a member than Mrs.Z will also not be a member so we will be left with (6+4-2) i.e. 8 people .So we need to select 3 from 8$8C3\$=56 ways.

Abcd
()

couldn't the explanation just be total combination 6! , but in half of them A will be before B and other half B will finish before A therefore answer is

6! / 2.