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If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.
There are 210 4 digit combinations and 5040 different 4 digit codes.
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
Assuming that the digits 0-9 are available, the answer is 5040. This is because for the first digit you can pick any of the original 10 digits. For the second you can have any otherdigit, so 9 possible choices. The third follows the same rule, so has 8 possibilities, and the forth digit, 7.This means that the total number of combinations is 10 x 9 x 8 x 7 = 5040
5040
10P4 = 5040
5040 numbers can be made.
If you can repeat the numbers within the combination there are 10,000 different combinations. If you cannot repeat the numbers within the combination, there are 5040 different combinations.
Well, darling, if you want a 4-digit combination with no repeated numbers using 0-9, you simply multiply the number of choices for each digit. So, for the first digit, you have 10 options, then 9 for the second (can't repeat the first digit), 8 for the third, and 7 for the fourth. Multiply those together and you get 10 x 9 x 8 x 7 = 5040 unique combinations. Hope that tickles your fancy!
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
Any 6 from 10 = 10 x 9 x 8 x 7 = 5040
This is a factorial problem. The first number can be any of ten digits, the second any of nine (because you can't repeat a digit), the third any of eight and the fourth any of the remaining 7 digits. 10x9x8x7=5040 combinations.