Question

A man X has 7 friends, 4 of them are ladies and 3 men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party is.
A.485
B.468
C.469
D.484

Answer 1

Hint: We will use the combination to find the number of ways of throwing the party. There are a number of ways in which guests can be invited, we will first calculate each probability then add them to find the total number of ways the party can be thrown.

Complete step-by-step answer:
First, we will make a table of the data given for easy understanding-

	Ladies	Men
X	4	3
Y	3	4

Case 1: - When 3 ladies friend of X is invited and 3 men friend of Y
$\Rightarrow$ Number of ways they can be invited are \[{{=}^{4}}C_{3}^{{}}{{\times }^{4}}C_{3}^{{}}=16\]

Case 2: - When 2 ladies from X and 1 lady from Y, then 1 man from X and 2 men from Y are invited.
$\Rightarrow$ Number of ways \[{{=}^{4}}C_{2}^{{}}{{\times }^{3}}C_{1}^{{}}{{\times }^{4}}C_{2}^{{}}=324\]

Case 3: - when 1 lady from X and 2 ladies from y, then from X and 1 man from Y
$\Rightarrow$ Number of ways \[{{=}^{4}}C_{1}^{{}}{{\times }^{3}}C_{2}^{{}}{{\times }^{3}}C_{2}^{{}}{{\times }^{4}}C_{1}^{{}}=144\]

Case 4: - when 0 ladies from X, that is 3 men from X and 3 ladies from Y
$\Rightarrow$ Number of ways \[ {{=}^{4}}C_{0}^{{}}{{\times }^{3}}C_{3}^{{}}=1\]
Now the total number of ways X and Y friend can invite to the party is \[=10+324+144+1=485\]

Note: Each possibility must be taken into account to find the number of ways to invite.
For easier understanding and keeping track of data sums should be divided into cases (in accordance with each probability).
A combination must be done to find a number of ways.