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

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485

468

469

484

answer is A.

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Detailed Solution

Case(I):When 3 ladies from X and 3 men from Y
No. of ways =

C_{3}^{4} \times C_{3}^{4}

=16
Case (ii): when 2 ladies from X and 1 lady from Y , then 1 man from X and 2 men from Y.
No. of ways =

C_{3}^{4} \times C_{1}^{3} \times C_{2}^{4}

=324
Case (iii) when 1 lady from X and 2 ladies from y, then from X and 1 man from Y
No. of ways =

C_{1}^{4} \times C_{2}^{3} \times C_{2}^{3} \times C_{1}^{4}

=144
Case (iv):when 0 lady from X i.e 3 men from X and 3 ladies from Y
No. of ways =

C_{3}^{3} \times C_{3}^{3}

=1
∴ Total no. of ways =16+324+144+1 =485

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