how many combinations can you get with 1 2 3 4 5?
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.
I should have also asked how many times the same number can be picked. I assumed only once.
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
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.
To find the number of combinations that can be formed using the numbers 1, 2, 3, 4, and 5, we can use the concept of permutation and combination. Since you have not specified the length of each combination, I will assume it is the same length as the number of elements given (i.e., a combination of all 5 numbers).
To determine the number of combinations, we will use the formula for combinations:
nCr = n! / (r! * (n-r)!)
Where n is the total number of elements and r is the length of each combination. In this case, n = 5 and r = 5.
So, plugging in the values, we have:
5C5 = 5! / (5! * (5-5)!)
Now, let's calculate the factorial terms:
5! = 5 * 4 * 3 * 2 * 1 = 120
5-5 = 0
Substituting these values back into the equation:
5C5 = 120 / (120 * 1)
Simplifying further:
5C5 = 1
Therefore, you can form only one combination using the numbers 1, 2, 3, 4, and 5, considering all 5 elements.