Question 957780: a committee of 5 people is to be chosen from 10 men and 12 women. How many ways can the committee be formed if it is to contain at least 2 women, AND one particular man (M1) and one particular woman(W1) DO NOT want to serve together?
I know at least two women will be (12C2)(20C3) and I think I need to subtract the number of committees that include both W1 and M1 from this number, OR add up the committees consisting of
(M1 alone) + (W1 alone) + (neither M1 or W1) but I'm stumped on how to proceed
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! looks like you have 2 problems here.
the first problem is you need to include at least 2 women.
this means you can include 2 women or 3 women or 4 women or 5 women on the committee since there is no requirement on the number of males that have to be on the committee.
the second problem is that you cannot have a particular woman paired with a particular man.
we'll tackle the first problem first and then we'll tackle the second problem.
there are 10 men and 12 women.
if 3 men and 2 women are on the committee, then the possible number of combinations is c((10,3) * c(12,2) which is equal to 7920.
if 2 men and 3 women are on the committee, then the possible number of combinations is c(10,2) * c(12,3) which is equal to 9900.
if 1 man and 4 women are on the committee, then the possible number of combinations is c(10,1) * c(12,4) which is equal to 4950.
if no men and 5 women are on the committee, then the possible number of combinations is c(10,0) * c(12,5) which is equal to 792.
you would need to add these up to get the total number of possible combinations.
7920 + 9900 + 4950 + 792 = 25362
now we'll tackle the second problem.
a particular women does not want to be with a particular man.
you would need to determine how many possible combinations this can happen in and subtract them from the total number of possible combinations.
if there are 2 women and 3 men on the committee, then this particular woman can be paired with 1 out of the 11 remaining women and this particular man can be paired with 2 of the 9 remaining men.
the total number of ways this can happen is c(11,1) * c(9,2) = 396.
with 3 women and 2 men, the total number of ways this can happen is c(11,2) * c(9,1) = 495.
with 4 women and 1 man, the total number of ways this can happen is c(11,3) * c(9,0) = 165.
with 5 women and 0 men, the total number of ways this can happen is c(11,4) * 0 = 0.
you need to add all these up and then subtract them from the total number of possible combinations to get your answer.
396 + 495 + 165 + 0 = 1056.
25362 - 1056 = 24306.
there are 24306 ways at least 2 women can be combined with men to form a committee of 5 people where one particular woman is not paired with one particular man.
that's my understanding of how this needs to be solved.
i worked it out with much smaller numbers and the method appears good.
c(n,x) is the combination formula of n! / (x! * (n-x)!)
for example:
c(9,2) is equal to 9! / (2! * 7!) which is equal to 36.
c(11,1) is equal to 11! / (1! * 10!) which is equal to 11.
11 * 36 = 396.
you saw this in the following statement copied from above:
the total number of ways this can happen is c(11,1) * c(9,2) = 396.
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