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Applying the formula n!/(n - r)! to the numbers n = 9 and r = 9, where n is the number of numbers to choose from, and r is the number of numbers chosen, 9! / (9 - 9)! is equal to 9 factorial. This is equal to 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880 combinations.

Q: How many combinations of 9 if you had to use all 9?

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9

9

10!/3! = 604800 different combinations.

10^9

There would be 9*9*9*9 or 6561 combinations.

Related questions

9

It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.

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.

9

10!/3! = 604800 different combinations.

10^9

There are: 9C6 = 84 combinations

I need to have listed all the 4 number combinations between 1 and 9

We can use 36 characters for each of the slots in the combination. Therefore, we have 36^11 possible combinations, or 131,621,703,842,267,136 combinations.

9

There would be 9*9*9*9 or 6561 combinations.

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