# Permutations and combinations

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• Sep 14th 2012, 08:35 AM
Shwems
Permutations and combinations
There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying)
• Sep 14th 2012, 08:49 AM
kalyanram
Re: Permutations and combinations
You have two offices (President, Treasurer). Now you have to choose 2 people from a group of 5 how many ways can this can be done.
Spoiler:
In $2.\binom{5}{2} = 2.\frac{5!}{2!3!} = 20$ ways. Multiplication by two is to allow permutations among them. Thanks to Evgeny (emakarov) for the correction.
• Sep 14th 2012, 09:10 AM
emakarov
Re: Permutations and combinations
Quote:

Originally Posted by Shwems
There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying)

See k-permutations.