SOLUTION: There are 9 candidates running for 3 seats on a committee. How many different election results are possible? (1 point) A)11 B)15 C)120 D)10

Algebra ->  Permutations -> SOLUTION: There are 9 candidates running for 3 seats on a committee. How many different election results are possible? (1 point) A)11 B)15 C)120 D)10       Log On


   

Question 876609: There are 9 candidates running for 3 seats on a committee. How many different election results are possible? (1 point)
A)11
B)15
C)120
D)10
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The 1st choice is 1 of 9
Then 1/8, then 1/7
--> 9*8*7 = 504
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But, ABC chosen is the same as ACB or CAB, 6 possible arrangements.
504/6 = 84 possible results.
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84
None of the above.