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Question
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’ s relatives?
Solution
Number of relatives of a man = 7
4 ladies and 3 gentlemen
Number of relatives of the man’s wife = 7
3 ladies and 4 gentlemen
The dinner party consists of 3 ladies and 3 gentlemen.
For the dinner party,
3 persons from the man’s relatives and 3 persons from the man’s wife’s relatives are invited.
Then we have the following possibilities for the different possible arrangements.
M | W | M | W | M | W | M | W | |
Ladies | 0 | 3 | 1 | 2 | 2 | 1 | 3 | 0 |
Gentlemen | 3 | 0 | 2 | 1 | 1 | 2 | 0 | 3 |
Required number of ways
= (4C0)(3C3) x (3C3)(4C0) + (4C1)(3C2)(3C2)(4C1) + (4C2)(3C1)(3C1)(4C2) + (4C3)(3C0)(3C0)(4C3)
= `1 xx 1 xx 1 xx 1 + 4 xx 3 xx 3 xx 4 + (4!)/(2!(4 - 2)!) xx (3!)/(1!(3 - 1)!) xx (3!)/(1!(3 - 1)!) xx (4!)/(2!(4 - 2)!) + ""^4"C"_1 xx 1 xx 1 xx ""^4"C"_1`
= `1 + 144 + (4 xx 3 xx 2!)/(2! xx 2!) xx (3 xx 2)/(2!) xx (3 xx 2!)/(2!) xx (4 xx 3 xx 2!)/(2! xx 2!) + 16`
= `145 + (4 xx 3)/(2 xx 1) xx 3 xx 3 xx (4 xx 3)/(2 xx 1) + 16`
= 145 + 2 × 3 × 3 × 3 × 2 × 3 + 16
= 145 + 324 + 16
= 485
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