How many combinations are there in a 9 digit number?

Answer 1

Does order matter? A common mistake is saying "combinations" when you actually mean "permutations". In the topic of Probability, permutations are ordered, while combinations are unordered.

This means that for permutations, "123456789" and "987654321" are each considered unique results. When dealing with combinations, they are considered the same result.

If you actually meant how many permutations are there, then the answer is as follows:

b9, where b is the base in which the number is expressed. In decimal, that would be 109, or 1000000000d. In binary, that would be 29, or 512d.

However, if you truly meant to ask how many combinationsexist for a 9 digit number, (assuming base 10 using digits 0 thru 9) the answer is: 10.

The formula for combinations is spoken as "n choose r", or nCr, where n is the number of available things to choose from, and r is how many things you are choosing from what is available. The combination formula is:

n! / (r! * (n - r)!)

So for your question, the equation would be as follows:

10! / (9! * (10-9)!) = 3628800 / (362880 * (1!)) = 3628800/362880 = 10

There are 10 combinations available for a 9 digit number when utilizing the digits 0 thru 9. Listing them out is as simple as excluding one of the digits from each combination:

123456789 (excludes 0)

012345678 (excludes 9)

012345679 (excludes 8)

012345689 (excludes 7)

012345789 (excludes 6)

012346789 (excludes 5)

012356789 (excludes 4)

012456789 (excludes 3)

013456789 (excludes 2)

023456789 (excludes 1)