JEE Mains · Maths · STD 11 - 6. permutation and combination
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 comman 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 \(484\)
- B \(485\)
- C \(468\)
- D \(469\)
Answer & Solution
Correct Answer
(B) \(485\)
Step-by-step Solution
Detailed explanation
\(X \to 4\) ladied \(Y \to 3\) \(X \to 3\) men \(Y \to 4\) Possible cases for \(X\) are \((1)\) \(3\) ladies, \(0\) man \((2)\) \(2\) ladies, \(1\) man \((3)\) \(1\) lady, \(2\) men \((4)\) \(0\) ladies, \(3\) men Possible cases for \(Y\) are \((1)\) \(0\) ladies, \(3\) men…
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