A committe of 5 is to be chosen from 6 men and 4 women, In how many ways can this be done if there must be more men than women in the committee
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The possibilities are for the committee are:
( 5 men and 0 women) OR ( 4 men and 1 woman) OR ( 3 men and 2 women)
Now we will determine in how many ways we could arrange each possibility.
( 5 men and 0 women) = 6*5 = 30 way
(4 men and 1 woman) = 6*4 + 1*4 = 24 + 4 = 28 ways
( 3 men and 2 women) = 6*3 + 2*4 = 18 + 12 = 30 way.
Now we will add all :
==> 30 + 28 + 30 = 88 way.
Then, there are 88 ways that we can create a committee of 6 men and 4 women.
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its too easy qustion:)
1stly,,,notice that condition says 6 mens nd 4 mens are there..
seconly..u need commeette of 5 people in which u need more men that women...simple
6c4*4c1+6c3*4c2...= 120....that's all there is no more steps to more men than women...
hope u all knw the above combination steps....
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