Permutations and Combinations.. In how many ways can a coach assign the 5 starting position in basketball to nine equally qualified men?
\[\frac{9!}{4!}\] ?
=15120
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
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
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
^^ ok
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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