Question

in how many ways can each women W1,W2,W3,W4 marry one of the six men M1,M2,M3,M4,M5,M6 under the following conditions. 1)W1 cannot marry M1 or M3 or M6 2)W2 cannot mary M2or M4​

Answer 1

Given : W1,W2,W3,W4 marry one of the six men M1,M2,M3,M4,M5,M6 under the following conditions. 1)W1 cannot marry M1 or M3 or M6 2)W2 cannot mary M2or M4​

To Find :  Number of Ways

Solution:

W1 cannot marry M1 or M3 or M6 hence can  marry to only M2 , M4 or M5

W2 cannot mary M2or M4​ hence can  marry to only M1 , M3 , M5 or M6

case 1 :

W1 Mary M2 or  M4   -  2 Ways

Then W2 can can marry in 4 ways  (M1 , M3 , M5 , M6)

Rest two women W3 and W4 can marry in ⁴P₂ = 4x3  = 12 Ways

2*4 * 12 = 96

case 2 :

W1 Mary M5   - 1 way

Then W2 can can marry in 3 ways  (M1 , M3  , M6)

Rest two women W3 and W4 can marry in ⁴P₂ = 4x3  = 12 Ways

1 * 3 * 12  = 36

96 + 36  = 132

in 132 Ways each women W1,W2,W3,W4 marry one of the six men M1,M2,M3,M4,M5,M6  if  1)W1 cannot marry M1 or M3 or M6 2)W2 cannot mary M2or M4​

if no conditions then ⁶P₄ =  360 ways

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