I think the number is 40,320.
Edit: Background…For work we need to create a unique code for fraud detection and the code can be no longer than 8 numerical digits. The generation of this number is another story. This is just for my own understanding.
In: 0
The way you find this is to ask “how many numbers can be in the first digit?”
9
Do this for all 8 digits. The math works out to 9 times 9 times 9…times 9 or 9^8
And that’s over 43,000,000. You ___ a few digits.
Your math includes all 1 2 3 4 5 6 and 7 digit numbers
You can make the numbers 1 through 40,320 and then some with only 5 digits, up to 99999.
8 digits is as you’d expect, 99999999
Combinations is the important bit here
You’re not saying “how many 8 digit numbers are there”, you’re being asked “how many different ways are there to arrange 8 different numbers”.
Change it to letters instead to make it clearer so you don’t think of it as a big number. You have the letters A, B, C, D, E, F, G, and H. How many different combinations can you make that have one and only one of each letter? ABCDEFGH is valid, so is BADCFEHG but AAAAHHHH is not valid because it isn’t a combination of all 8
For the first spot you can pick from any of the 8, for the second slot you can only pick from the *remaining* 7 letters, the third slot is 6 *remaining*, and so on. The final number of combinations is 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40,320
The number of unique combinations using 8 numbers is 99999999 but I believe what you’re talking about is the number of combinations of 8 digits where each digit is used only once. And that’s why it’s less than 99999999. 12345678 is valid but 11111111 is not, 11111112 is not, etc. so you see it’s waaay less than 99999999
Simplify it to two numbers. That answer isn’t 99, it’s 2, which is equal to 2!(two factorial, which is 2 times 1). Then three numbers, it’s not 999, it’s 3!, 3x2x1 which is 6. Four unique digits (or characters) can be arranged 4! ways. In the first place we have 4 uniques to pick from, then in the second place we have 3 uniques, third place we have 2 to pick from and the last place leaves us one option. That’s 4x3x2x1, or 4! or 24.
99999999 is the correct answer if you include 0 as a number that can be used. For example if the number ’20’ is written as an 8 digit number it’s ‘00000020’. This gets included in your example because it’s not a string of all 0’s.
0-9 gives you 10 possible numbers to use.
If they can be used multiple times, then for 8 digits you have 10^8 combinations. 10 choices for the first digit, 10 for the second… all the way to the 8th digit.
That gives you 100000000 combinations, if you exclude ‘00000000’ then you are left with 99999999 combinations.
What you are looking for is actually permutations, or number of ways of arranging something.
Combinations are something different in math. In a combination order doesn’t matter. So 12345678 and 87654321 are considered the same combination of numbers. You see this in the LOTTO. it doesn’t matter which order the numbers are drawn only that you match the numbers chosen.
Permutations are ways of arranging things in order. So if you had say A B and C how many ways can you order them? ABC, ACB, BAC, BCA, CAB, CBA. Six, there are six permutations of three letters (or numbers or colors, etc.). For N unique objects (where N is the number of objects) you can calculate this by using the simple formula of N! (said N factorial). N! just means to multiple like this: N * N-1 * N-2 * N-3 *… * 2 * 1. For 3 this is just 3 * 2 * 1 = 6, just like we saw above. For 8 this is 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40,320.
Latest Answers