# math

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how many combinations can you get with 1 2 3 4 5?

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Combinations 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.

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I should have also asked how many times the same number can be picked. I assumed only once.

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I 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 -

If 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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