To calculate the number of combinations of four numbers from a set of 32 numbers, you can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 32 (total numbers) and r = 4 (numbers chosen). Plugging these values into the formula, you get 32C4 = 32! / (4!(32-4)!) = 32! / (4!28!). After calculating this expression, you will find that there are 35,960 possible combinations of four numbers from a set of 32 numbers.
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.
Assuming there are 5 check boxes and the respondent may check any number of them, the answer is 25 = 32 combinations.
-30
There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.
there are altogether 24 numbers between 9 and 32. by using nCr, 24C4 = 10626
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.
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
There are 35C4 = 35*34*33*32/(4*3*2*1) = 52,360 combinations.
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 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
If repetitions are allows, 325. If no repetitions are allows, then it is 32 x 31 x 30 x 29 x 28.
( 36! )/( 32! ) ( 4! ) = (36 x 35 x 34 x 33)/(4 x 3 x 2) = 58,905
It depends on how many numbers make one combination.Here are the possibilities for combinations of several different sizes:Numbers in the combinationPossible combinations1 . . . . . 322 . . . . . 4963 . . . . . 4,9604 . . . . . 35,9605 . . . . . 194,4326 . . . . . 906,1927 . . . . . 3,365,8568 . . . . . 10,518,300
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
32 of them.