A man has 7 relatives (4 women and 3 men). His wife also has 7 relatives (3 women and 4 men). In how many ways can they invite 3 women and 3 men so that 3 of them are man's relatives and 3 of them are his wife's relatives?

  1. A. 340
  2. B. 484
  3. C. 485 ✓
  4. D. 469

Correct Answer: C. 485

Explanation

The required team has 3 women and 3 men. If the husband invites x women and y men, the wife must invite 3-x women and 3-y men. Also x+y=3. The possible pairs of (x,y) are (3,0), (2,1), (1,2), (0,3). Number of ways = \binom{4}{3}\binom{3}{0}\binom{3}{0}\binom{4}{3} + \binom{4}{2}\binom{3}{1}\binom{3}{1}\binom{4}{2} + \binom{4}{1}\binom{3}{2}\binom{3}{2}\binom{4}{1} + \binom{4}{0}\binom{3}{3}\binom{3}{3}\binom{4}{0} = 16 + 324 + 144 + 1 = 485.

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