Question 918991: Having a lot of trouble with understanding Probabilities, Combinations, and Permutations. Please help. When I looked up a problem similar to this one I saw something like ! by the numbers (5! + 4! = 126) which was pretty confusing.
Exact Question: How many different 5-person basketball teams can a coach choose from among 7 players?
I thought all you had to do was multiply 5 x 7 which gives you 35, but I'm doubting the answer.
Thank you so so very much for reading this question and if you do proceed to answer please answer which ALOT detail. Sorry if I sound demanding, I can't stress how much probability, combination, and premutation confuses me. >.<
Answer by Hawksfan(61)   (Show Source):
You can put this solution on YOUR website! In this problem you will either have a permutation or a combination. But how do you tell?
With a permutation, the order or how the items are arranged matters greatly.
With a combination, the order doesn't matter at all.
The example above talks about a problem where it doesn't matter what
order the players are in. We can have the 5 player teams in any order.
Therefore its a combination.
The formula for a combination is nCr = n!/(n-r)!
where n = total number of items
r = number of items of subgroup
examples of n! (separate from the problem)
4! = 4*3*2*1
9! = 9*8*7*6*5*4*3*2*1
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nCr = n!/r!(n-r)!
n = 7, r =5
7C5 = 7!/5!(7-5)! = 7*6*5*4*3*2*1/[(5*4*3*2*1)(2*1)]
7C5 = 7*6/2
= 21
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