Sorry for the straightforward question. Yet if we have actually a 5-digit number and also each digit have the right to hold a worth from 0-9. Then just how many possible combinations space there? Also, what formula deserve to you usage to gain this in other cases of different digits. (If any type of at all).

You are watching: How many number combinations in 9 digits In general, in basic b, there would be bn feasible combinations of n digits. For your question, you deserve to substitute b = 10 and n = 5 to gain 100000. You might have accidentally excluded the combination 00000 once you came up through 99999 as the number of combinations.

However, this consists of combinations that start with 0 (e.g. 00135), which space not normally taken into consideration to it is in five-digit numbers. If you want to exclude those (and presume n > 1), then you subtract away every the combinations starting with 0, of i beg your pardon there space bn−1, to obtain bn−bn−1 = (b−1)bn−1 together the last answer.

Just to do this much less abstract and an ext concrete:

If you exclude 0 because that the very first digit, you have 9 alternatives for it: 1, 2, 3, 4, 5, 6, 7, 8, 9. For the other digits, you have 10 options to choose from.

So the total variety of combinations is 9×10×10×10×10 = 9×104 = (b-1) bn-1 because that b = 10. The number of 5-digit combinations is 105=100,000. So, one more than 99,999. You have the right to generalize that: the variety of N-digit combinations is 10N.

Now, this assumes that you count 00000 or 00534 as "5-digit numbers". If you pick to no count any combination of 5 digits starting with a leading 0, then the very first 5-digit number is 10,000 and the critical 5-digit number is 99,999. That gives you 90,000 5-digit numbers. In that instance the basic formula would be 9*10N-1 because that the variety of possible N-digit numbers.

You have the right to generalize this formula to other bases as well. Since there room 26 letters and also 10 1-digit numbers, then a case-sensitive alpha-numeric password includes 26+26+10=62 possible inputs because that each character. An N-character password would because of this have 62N possible combinations.

Assuming the very first digit can be 0, there room 100000 such combinations (because after ~ chopping off the early zeroes, each mix gives a number between 0 and 99999 and also there space obviously 100000 that those).

If the first digit can't be zero (so the the number is genuinely 5 digits long), there are 100000 - 10000 = 90000 (we are simply taking far 00000 v 09999).

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Maybe the simplest method is just to count, no permutations/comb necessary. The smallest 5 digit number is 10000, biggest 99999. And also then simply their distinction is 89999, then add 1 to encompass the very first number, therefore 90000.