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

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Q: How many 3 digit combinations numbers one through nine?

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Exactly 3,628,800, or 10!.

10!/3! = 604800 different combinations.

If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.

There are 360 of them.

That's exactly all the numbers you need to count from 000,000,000 to 999,999,999.There are one billion of them.

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Number of 7 digit combinations out of the 10 one-digit numbers = 120.

There are 38760 combinations.

Exactly 3,628,800, or 10!.

The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.

It is: 9C7 = 36

66

15

10,000

2304

There are 1140 five digit combinations between numbers 1 and 20.

10!/3! = 604800 different combinations.

that would be 9 to the 7th power i believe.

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