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Mathematics 7 Online
OpenStudy (anonymous):

Permutations and Combinations.. In how many ways can a coach assign the 5 starting position in basketball to nine equally qualified men?

OpenStudy (anonymous):

\[\frac{9!}{4!}\] ?

OpenStudy (anonymous):

=15120

OpenStudy (anonymous):

This is a good question. Does the order of choosing matter since each position on a basketball court is technically different? i over thought these question so bad in high school lol

OpenStudy (anonymous):

In real life yep, but in this problem team members are "equally qualified" so no. Besides there are not enough information. Thus : For position 1, coach has 9 possibilities For position 2, 8 possibilities ... This leads to 9*8*7*6*5 = 15120

OpenStudy (anonymous):

ohh yes i completely agree with your answer. I'm just saying the thoughts that go through my head which i must suppress. ha ha

OpenStudy (anonymous):

^^ ok

OpenStudy (chaise):

If you search "patrickJMT - combinations and permutations" you will get some nice little videos on permutations and combinations, with really good explanations. If you're having troubles or are curious :)

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