Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
120
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
There are 5,040 combinations.
There are 7 digits from 0-6 To determine if a number is divisible by 4, we only need look at the last 2 digits. Possibilities are: [135][26] => 12, 16, 32, 36, 52, 56 and [0246][04] => 04, 20, 24, 40, 60, 64 (ignoring 00 and 44) There are 8 combinations without a 0, and 4 with a 0. Of the 8 combinations without a 0, We can only pick 4 of the remaining digits for the ten thousand digit. This leaves us with 4 digits for the thousands digit, 3 for the hundreds digit Of the 4 combinations with a 0, similarly we have 5 X 4 X 3 possibilities for the rest of the digits. 8 * 4 * 4* 3 + 4 * 5 * 4 * 3 = 32 * 12 + 20 * 12 = 52 * 12 = 624 There are 624 possible numbers.
Since there are only five different digits, a 6-digit number can only be generated if a digit can be repeated. If digits can be repeated, the smallest 6-digit number is 111111.
If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as: nPr = n! / (n-r)! 10P4 = 10! / (10-4)! 10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6 10P4 = 7 * 8 * 9 * 10 10P4 = 5040 There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
It can be calculated as factorial 44! = 4x3x2x1= 60
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
128
1
There are 5,040 combinations.
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
There are 210 4 digit combinations and 5040 different 4 digit codes.
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
As the number has to start with 15, we have only 3 remaining digits to work with. There are 3 possible options for the first digit. Then out of each of these, 2 possible options for the second digit, and one option for the last. This means that in total there are 3x2x1 (6) possible combinations. These are: 15234 15243 15324 15342 15423 15432