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Q: How many combinations are there of 8 numbers?

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10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210

If each can be used again, it is 7x7x7x7 or 2401 combinations. If not it is 7x6x5x4 or 840 combinations

There are 8.There are 8.There are 8.There are 8.

Infinitely many. 1 + 1 + 1 + 19 .1 + .1 + .1 + 21.7 .01 + .01 +.01 + 21.97 And so on. And all these are with three of the numbers being the same. There are also combinations where some numbers are irrational -infinite, non recurring decimals. Then there are combinations with some of the numbers being negative.

11

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There are 125970 combinations and I am not stupid enough to try and list them!

There are 8*7/(2*1) = 28 combinations.

two

8C4 = 70

8C3 = 56 of them

64

41

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.

56 combinations. :)

nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.

There are 8!/[6!(8-6)!] = 8*7/2 = 28 - too many to list.

how many combinations of 4 numbers are there in 7 numbers

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