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).

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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.