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
(c) 469
(d) 484
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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.
$\therefore$ 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 \\ & \quad+{ }^{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} $
