Q: How many permutations of the numbers 10 through 14 are there?

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Since there are 10 numbers to fill 4 spots, each spot would have 10 choices for which number would fill it. We have (10)(10)(10)(10)=1000.

10! permutations of the word "Arithmetic" may be made.

Not counting negative numbers (which would give you an infinite number of ways of adding numbers together to equal 10), there are 6 different ways (permutations) of adding numbers together to equal 10. 0 + 10 1 + 9 2 + 8 3 + 7 4 + 6 5 + 5

90

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.

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Since there are 10 numbers to fill 4 spots, each spot would have 10 choices for which number would fill it. We have (10)(10)(10)(10)=1000.

10! permutations of the word "Arithmetic" may be made.

There are 75600 permutations.

In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.

2

Not counting negative numbers (which would give you an infinite number of ways of adding numbers together to equal 10), there are 6 different ways (permutations) of adding numbers together to equal 10. 0 + 10 1 + 9 2 + 8 3 + 7 4 + 6 5 + 5

90

The answer is 20C10 which is 20!/[10!(20-10)!] = 184756

If all numbers can be used as many times as wanted then there are 109 = one billion combinations. If each number can be used only once, there are 10!/(10-9)! = 10!/1! = 10! = 3628800 combinations. * * * * * Clearly answered by someone who does not know the difference between PERMUTATIONS and COMBINATIONS. The combination 123456789 is the same as the combination 213456789 etc. All in all, therefore, there are only ten combinations which use each digit at most once.

There are 720 of them.

ABC A = 1,2,3,4,5,6,7,8,9; 9 possibilities B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities So there are 9*10*10 = 900 numbers with 3 digits.

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.