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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?









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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)


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


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.


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.