Question

# A man X has 7 friends, 4 of them are ladies and 3 are 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?

Hint: In this question combination formula will be used for getting a suitable group. There will be possibly four cases consisting of different varieties of groups of men and ladies. Final answer will be the total of all four cases.

Here, Friends of X = 4 Ladies + 3 Men
Friends of Y = 3 Ladies + 4 Men
Case-1: When 3 ladies from X and 3 men from Y will be invited.
So, no. of ways =
$C(4,3) \times C(4,3) \\ = 4 \times 4 \\ = 16 \\$
Case-2: When 2 ladies from X and 1 lady from Y will be invited with 1 man from X and 2 men from Y.
So, no. of ways =
$C(4,2) \times C(3,1) \times C(3,1) \times C(4,2) \\ = 6 \times 3 \times 3 \times 6 \\ = 324 \\$
Case-3: When 1 lady from X and 2 ladies from Y, and from 2 men from X and 1 man from Y will be invited.
So, no. of ways =
$C(4,1) \times C(3,2) \times C(3,2) \times C(4,1) \\ = 4 \times 3 \times 3 \times 4 \\ = 144 \\$
Case-4: When 0 lady from X means 3 men from X and 3 ladies from Y will be invited.
So, no. of ways =
$C(3,3) \times C(3,3) \\ = 1 \times 1 \\ = 1 \\$
$\therefore$ Total number of ways will be = 16 + 324 + 144 + 1
= 485

Note: Probability is the possible number of outcomes to the total number of outcomes. Permutations is the arrangement of the items in the set. Combinations is the selection of items from the total number of items in a set.