Question 166938: How many differant committees consisting of 3 teenagers and 5 youths are possible if each member of the committee is to be chosen from a group of 6 teenagers and 10 youths
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! If the order doesn't matter, it is a Combination.
If the order does matter it is a Permutation.
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In your case, "order" does not matter so it would be a combination.
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C(n,r) = n!/(r!(n-r)!)
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3 teenagers from a group of 6 teenagers
n=6
r=3
C(n,r) = n!/(r!(n-r)!)
C(6,3) = 6!/(3!(6-3)!)
C(6,3) = 6!/(3!(3)!)
C(6,3) = (1*2*3*4*5*6)/(3!(1*2*3))
C(6,3) = (4*5*6)/(3!)
C(6,3) = (4*5*6)/(1*2*3)
C(6,3) = 120/6
C(6,3) = 20
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5 youths from a grou of 10 youths
n=10
r=5
C(n,r) = n!/(r!(n-r)!)
C(10,5) = 10!/(5!(10-5)!)
C(10,5) = 10!/(5!(5)!)
C(10,5) = (6*7*8*9*10)/(5!)
C(10,5) = (6*7*8*9*10)/(1*2*3*4*5)
C(10,5) = 30240/120
C(10,5) = 252
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Possible committees:
20*252 = 5040
How many differant committees consisting of 3 teenagers and 5 youths are possible if each member of the committee is to be chosen from a group of 6 teenagers and 10 youths
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