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.
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.
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
but
I suspect you mean permutations.
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.