There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations.
I do not have the inclination to list them all.
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.
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
Two . . . . . 38 and 83.
Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.
There are 14C5 = 14*13*12*11*10/(5*4*3*2*1) = 2002
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.
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
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
9
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
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.
8C3 = 56 of them
There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
I dont understand what you are asking.
There are: 10C3 = 120
13 combinations of 3