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This is a factorial problem. The first number can be any of ten digits, the second any of nine (because you can't repeat a digit), the third any of eight and the fourth any of the remaining 7 digits. 10x9x8x7=5040 combinations.

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Q: If you create a password using 4 numbers 0 through 9 how many different combinations are there without using the same number twice?

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10!/3! = 604800 different combinations.

575757

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.

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.

Related questions

1000

10!/3! = 604800 different combinations.

66

15

10,000

575757

2304

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.

Through the magic of perms and coms the answer is 729

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.

There are 3,628,800 different permutations that can be made when using the numbers 0 through 9, if each number is only used once in each combination.

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