Sorry because that the basic question. However if we have a 5-digit number and each digit have the right to hold a value from 0-9. Then exactly how many possible combinations space there? Also, what formula deserve to you usage to gain this in other situations of various digits. (If any type of at all).
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In general, in base b, there would certainly be bn possible combinations the n digits. For your question, you have the right to substitute b = 10 and n = 5 to obtain 100000. You could have accidentally excluded the mix 00000 as soon as you came up with 99999 as the number of combinations.
However, this consists of combinations that begin with 0 (e.g. 00135), which space not normally taken into consideration to it is in five-digit numbers. If you desire 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 get bn−bn−1 = (b−1)bn−1 as the last answer.
Just to do this much less abstract and much more concrete:
If you exclude 0 because that the first digit, you have 9 choices for it: 1, 2, 3, 4, 5, 6, 7, 8, 9. Because that the various other digits, you have actually 10 alternatives to choose from.
So the total number of combinations is 9×10×10×10×10 = 9×104 = (b-1) bn-1 because that b = 10.
The number of 5-digit combine is 105=100,000. So, one more than 99,999. You deserve to generalize that: the number of N-digit combine is 10N.
Now, this assumes the you counting 00000 or 00534 together "5-digit numbers". If you pick to not count any combination of 5 digits beginning with a leading 0, then the first 5-digit number is 10,000 and also the last 5-digit number is 99,999. That provides you 90,000 5-digit numbers. In that instance the basic formula would be 9*10N-1 because that the number of possible N-digit numbers.
You deserve to generalize this formula to various other bases as well. Because there space 26 letters and also 10 1-digit numbers, climate a case-sensitive alpha-numeric password contains 26+26+10=62 possible inputs because that each character. One N-character password would therefore have 62N possible combinations.
Assuming the an initial digit deserve to be 0, there room 100000 together combinations (because after ~ chopping turn off the initial zeroes, each combination gives a number in between 0 and also 99999 and there room obviously 100000 that those).
If the very first digit can't be zero (so the the number is genuinely five digits long), there room 100000 - 10000 = 90000 (we are simply taking away 00000 v 09999).
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Maybe the simplest means is simply to count, no permutations/comb necessary. The smallest 5 number number is 10000, biggest 99999. And also then simply their difference is 89999, then include 1 to encompass the first number, so 90000.