Q: How many combinations of numbers can you make with 8 and 3?

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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.

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 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.

There are 14C5 = 14*13*12*11*10/(5*4*3*2*1) = 2002

Related questions

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.

You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.

13 combinations of 3

14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.

None. You do not have enough numbers to make even one combination.

9

The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.

There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.

You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.

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.