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March 27, 2015

March 27, 2015

Posted by **shirley** on Sunday, November 30, 2008 at 8:06pm.

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**drwls**, Sunday, November 30, 2008 at 8:11pmCombinations of how many numbers? Don't you mean permutations?

If 1,2,3,4 and 5 are your only five numbers and you must pick five, and if the order does not matter, then there is only one combination.

- math -
**drwls**, Sunday, November 30, 2008 at 8:13pmI should have also asked how many times the same number can be picked. I assumed only once.

- math -
**Damon**, Sunday, November 30, 2008 at 8:16pmI assume that all of the elements are included in each combination (you can not use 132 ignoring 4 and 5 for example)

In that case the question is how many arrangements can be made of n things taken n at a time which is the permutations of n elements taken r at a time where here r = n = 5

That implies that each element is used only once

0! = 1 by the way

P(n,r) = n!/(n-r)!

P (5,5) = 5!/0! = 5*4*3*2*1 =20*6 = 120

In other words I have 5 choices for the first number

four choices for the second number

3 choices for the third

2 for the fourth

and only one for the last

- math -
**Damon**, Sunday, November 30, 2008 at 8:19pmIf you really mean "combinations" then Dr WLS has it right. If there is no matter the order (combinations) then there is only one combination

but

I suspect you mean permutations.

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