Question 1203955: A committee of 6 U.S. senators is to be formed with 3 Democrats and 3 Republicans. In how many ways can this be done if there are 56 Democratic senators and 44 Republican senators?
Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20850)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
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There are 56*55*54 = 166,320 permutations of selecting 3 Democrats from a pool of 56.
We start at 56 and count down until we have filled 3 slots. Multiply the values along the way.
On a committee, no member outranks another. There aren't special named seats such as "president", "VP", "treasurer", etc.
The lack of named seats indicates that order doesn't matter.
A committee like {Alice, Bob, Carol} is the same as {Carol, Alice, Bob}.
Because order doesn't matter on a committee, we divide by 3! = 3*2*1 = 6.
There are 6 ways to rearrange {a,b,c}. We must divide by 6 to correct for the erroneous overcount.
(166,320)/6 = 27,720
There are 27,720 different combinations of 3 Democrats from a pool of 56.
Another way to arrive at the value 27,720 is to use the nCr combination formula with n = 56 and r = 3
n C r = (n!)/(r!(n-r)!)
56 C 3 = (56!)/(3!*(56-3)!)
56 C 3 = (56!)/(3!*53!)
56 C 3 = (56*55*54*53!)/(3!*53!)
56 C 3 = (56*55*54)/(3!)
56 C 3 = (56*55*54)/(3*2*1)
56 C 3 = 166320/6
56 C 3 = 27720
Pay careful attention to the fact, on the third to last step, we have "56*54*53" in the numerator and "3*2*1" in the denominator.
Through similar calculations we have 44C3 = 13,244 different trios of Republicans where we have select from a pool of 44.
Overall there are 27720*13244 = 367,123,680 different committees where order doesn't matter.
Answer: 367,123,680
(approximately 367 million)
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