SOLUTION: A husband has 7 relatives, 4 of them are women and the rest men. His wife too has 7 relatives out of which 3 are women and 4 men. In how many ways can they invite a dinner par

Algebra ->  Permutations -> SOLUTION: A husband has 7 relatives, 4 of them are women and the rest men. His wife too has 7 relatives out of which 3 are women and 4 men. In how many ways can they invite a dinner par      Log On


   

Question 943735: A husband has 7 relatives, 4 of them are
women and the rest men. His wife too
has 7 relatives out of which 3 are
women and 4 men.
In how many ways can they invite a
dinner party of 3 ladies and 3 men so
that there are 3 of man's relatives and
3 of wife's ?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
There are 4 cases:

         Husband's  |    Wife's
         relatives  |   relatives
Case      invited   }    invited
----------------------------------
 1.       3M, 0W    |    0M, 3W
 2.       2M, 1W    |    1M, 2W
 3.       1M, 2W    |    2M, 1W
 4.       0M, 3W    |    3M, 0W

Case 1: Choose the husband's 3 men relatives 3C3 = 1 way.
        Choose the wife's 3 women relatives 3C3=1 ways.
        That's 1*1 = 1 way for Case 1.

Case 2: Choose the husband's 2 men relatives 3C2 = 3 ways.
        Choose the husband's 1 woman relative 4C1 = 4 ways.
        Choose the wife's 1 man relative 4C1=4 ways.
        Choose the wife's 2 women relatives 3C2 = 3 ways.
        That's 3*4*4*3 = 144 ways for Case 2.

Case 3: Choose the husband's 1 man relative 3C1 = 3 ways.
        Choose the husband's 2 women relatives 4C2 = 6 ways.
        Choose the wife's 2 men relative 4C2 = 6 ways.
        Choose the wife's 1 woman relative 3C1 = 3 ways.
        That's 3*6*6*3 = 324 ways for Case 3.

Case 4: Choose the husband's 3 women relatives 4C3 = 4 ways.
        Choose the wife's 3 men relatives 4C3=4 way.
        That's 4*4 = 16 ways for Case 4.

Grand total from all cases:  1+144+324+16 = 485 ways.

Edwin