The answer is 32!/(27! * 5!) where n! represents 1*2*3*... *n
So the answer is 32*31*30*29*28/(5*4*3*2*1)
= 201 376
5*4*3*2*1 = 120 combinations. * * * * * No. The previous answerer has confused permutations and combinations. There are only 25 = 32 combinations including the null combinations. There is 1 combination of all 5 numbers There are 5 combinations of 4 numbers out of 5 There are 10 combinations of 3 numbers out of 5 There are 10 combinations of 2 numbers out of 5 There are 5 combinations of 1 numbers out of 5 There is 1 combination of no 5 numbers 32 in all, or 31 if you want to disallow the null combination.
120
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
32 of them.
31C5 = 169911
There are: 17C5 = 6188
5
There are 252 combinations.
120 WRONG! That is the number of PERMUTATIONS. In the case of combinations, the order of the numbers does not matter, so there is only 1 5-number combination from 5 numbers.
41