Sorry because that the an easy question. Yet if we have actually a 5-digit number and also each digit can hold a value from 0-9. Then just how many feasible combinations space there? Also, what formula have the right to you usage to obtain this in other instances of different digits. (If any type of at all).
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In general, in base b, there would be bn possible combinations of n digits. For your question, you can substitute b = 10 and n = 5 to acquire 100000. You can have accidentally excluded the combination 00000 as soon as you came up v 99999 as the variety of combinations.
However, this contains combinations that begin with 0 (e.g. 00135), which room not normally thought about to be five-digit numbers. If you want to exclude, those (and suspect n > 1), then you subtract away every the combinations starting with 0, of i m sorry there are bn−1, to acquire bn−bn−1 = (b−1)bn−1 together the final answer.
Just to make this much less abstract and an ext concrete:
If girlfriend exclude 0 for the first digit, you have actually 9 alternatives for it: 1, 2, 3, 4, 5, 6, 7, 8, 9. Because that the other digits, you have 10 options to pick from.
So the total number of combinations is 9×10×10×10×10 = 9×104 = (b-1) bn-1 because that b = 10.
The variety of 5-digit combine is 105=100,000. So, one an ext than 99,999. You have the right to generalize that: the number of N-digit combine is 10N.
Now, this assumes that you count 00000 or 00534 together "5-digit numbers". If you choose to not count any combination of 5 digits beginning with a top 0, then the very first 5-digit number is 10,000 and the last 5-digit number is 99,999. That provides you 90,000 5-digit numbers. In that case the general formula would certainly be 9*10N-1 because that the variety of possible N-digit numbers.
You can generalize this formula to other bases as well. Because there are 26 letters and 10 1-digit numbers, climate a case-sensitive alpha-numeric password consists of 26+26+10=62 feasible inputs for each character. An N-character password would as such have 62N feasible combinations.
Assuming the very first digit can be 0, there are 100000 together combinations (because ~ chopping turn off the early stage zeroes, each mix gives a number between 0 and also 99999 and there room obviously 100000 of those).
If the an initial digit can't be zero (so that the number is genuinely 5 digits long), there room 100000 - 10000 = 90000 (we are simply taking far 00000 through 09999).
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Maybe the simplest way is simply to count, no permutations/comb necessary. The smallest 5 digit number is 10000, biggest 99999. And then simply their difference is 89999, then include 1 to incorporate the first number, so 90000.