Method 1: four digit number start after 999 and end at 9999
so 4 digit no. = 9999-999 = 9000
method 2: - - - - there are four places in 4 digit no. In first place only 1-9 can come , for rest 0-10
so total no. = 9*10*10*10=9000
* * * * *
Wrong!
Those are the number of PERMUTATIONS, not combinations. In a combination, the order of digits does not matter. So 1234 is the same as 1342 or 4213 etc.
There are more than 4.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
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.
For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.
Too many to list here-see below.
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
There are more than 4.
I can't
There are 840 4-digit combinations without repeating any digit in the combinations.
There are 210 4 digit combinations and 5040 different 4 digit codes.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
16
i would like a list all possible 4 digit combination using 0-9
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There would be 9*9*9*9 or 6561 combinations.
There are 5,040 combinations.
Passwords are technically permutations, not combinations. There are 104 = 10000 of them and I regret that I do not have the time to list them. They are all the numbers from 0000 to 9999.