SATHEE: Permutations and Combinations 2 Question 8

Permutations and Combinations 2 Question 8

8. 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

(2017 Main)

(a) 485

(b) 468

(d) 484

Show Answer

Answer:

Correct Answer: 8. (a)

Solution:

Given, $X$ has 7 friends, 4 of them are ladies and 3 are men while $Y$ has 7 friends, 3 of them are ladies and 4 are men.

$∴$ Total number of required ways

$\begin{aligned} =^{3} C_{3} \times^{4} C_{0} \times^{4} C_{0} \times^{3} C_{3} +^{3} C_{2} \times^{4} C_{1} \times^{4} C_{1} \times^{3} C_{2} \\ +^{3} C_{1} \times^{4} C_{2} \times^{4} C_{2} \times^{3} C_{1} +^{3} C_{0} \times^{4} C_{3} \times^{4} C_{3} \times^{3} C_{0} \\ = 1 + 144 + 324 + 16 = 485 \end{aligned}$

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Permutations and Combinations 2 Question 8

8. 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

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8. 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