From A Total Of Six Men And Four Ladies A Committee Of Three Is To Be

1.	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?
Answer» 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. 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.

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?

Answer»

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.