I believe that the problem should read as follows:
There are 11 women and 9 men in a certain club. If the club is to select a committee of 2 women and 2 men, how many different such committees are possible?
A. 120
B. 720
C. 1,060
D. 1,520
E. 1,980
Women:
Number of options for the first woman = 11. (Any of the 11 women.)
Number of options for the second woman = 10. (Any of the 10 remaining women.)
To combine these options, we multiply:
11*10.
Since the ORDER of the women does not matter -- AB constitutes the same pair of women as BA -- we DIVIDE by the number of ways the two women can be ARRANGED (2!):
(11*10)/(2*1) = 55.
Men:
Number of options for the first man = 9. (Any of the 9 men.)
Number of options for the second man = 8. (Any of the 8 remaining men.)
To combine these options, we multiply:
9*8.
Since the ORDER of the men does not matter -- CD constitutes the same pair of men as DC -- we DIVIDE by the number of ways the two men can be ARRANGED (2!):
(9*8)/(2*1) = 36.
To combine our 55 options for the 2 women and our 36 options for the 2 men, we MULTIPLY:
55*36 = 1980.
The correct answer is
E.
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