Question 849795: i really don't understand Cominations..... so here is my question: In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have 8 men and 10 women and you are trying to draw a committee that consists of 5 men and 6 women.
the number of ways you can get 5 men out of 8 men is equal to 8C5 which is equal to 56
the number of ways you can get 6 women out of 10 women is equal to 10C6 which is equal to 210
the total number of ways you can get a committee of 5 men and 6 women where each committee has at least one member that is different from any other committee is equal to 56 * 210 = 11760
with all those combinations, its impossible to show you how this works.
i can, however, show you with a much simpler example.
assume you want to draw a committee of 2 men and 2 women from a group that consists of 3 men and 3 women.
we'll use the same formulas to see what the answer would be and then we'll show you how this works.
number of possible ways to get 2 men out of 3 is equal to 3C2 which is equal to 3.
number of possible ways to get 2 women out of 3 is equal to 3C2 which is equal to 3.
total number of possible committees that can be formed where are least one of the members is not the same as any other committee is equal to 3 * 3 = 9.
now let's see how this looks.
let the men be equal to a,b,c
let the women be equal to 1,2,3
the number of possible ways to get 3 men is 3.
those ways are:
a,b
a,c
b,c
the number of possible ways to get 3 women is 3.
those ways are:
1,2
1,3
2,3
the 9 possible committees consisting of 2 men and 2 women each are:
a,b,1,2
a,b,1,3
a,b,2,3
a,c,1,2
a,c,1,3
a,c,2,3
b,c,1,2
b,c,1,3
b,c,2,3
the combination formula of 8C5 is equal to 8! / (5! * 3!)
the combination formula of 10C6 is equal to 10! / (6! * 4!)
the combination formula of 3C2 is equal to 3! / (2! * 1!)
you are forming groups where order is not important.
that's the combination formula.
if order was important, then you would use the permutation formula.
no replacement is assumed.
once you draw one person, you go back and draw another person without replacing the first person back in the group.
here's a reference that might help.
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
there are others.
just do a search on "combinations and permutations" and you'll get a whole list of them.
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