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Permutation problem

by umaa » Wed Jul 29, 2009 2:41 pm
In a race C, seedings are determined as follows: seeds 1, 2, and
3 are given to the top three …finishers in race A, while seeds 4, 5,
and 6 are given to the top three finishers in race B. If race A and
race B both consist of six entrants, how many possible
arrangements of seeds are there for race C?
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by jjk » Wed Jul 29, 2009 3:34 pm
For Race A, there are 6 entrants and 3 spots.

6 entrants can take the first spot, leaving 5 to take the second, and 4 to take the third. 6 X 5 X 4 = 120

For Race B, the same calculation method applies here.

6 entrants can take the first spot, leaving 5 to take the second, and 4 to take the third. 6 X 5 X 4 = 120

120 possible outcomes for Race A and 120 possible outcomes for Race B, so 120 X 120 = 14,400.
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by umaa » Wed Jul 29, 2009 3:54 pm
Thanks.. Thats the answer.
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