24 without repeating digits: 1938 1983 1398 1389 1893 1839 9138 9183 9318 9381 9813 9831 3198 3189 3918 3981 3819 3891 8193 8139 8913 8931 8319 and 8391.
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
45*44*43*42*41/(5*4*3*2*1) = 1,221,759
9999 + 1 * * * * * That is so wrong! There are 10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers from the set of ten 1-digit numbers, 0 1, 2, ... 9 When considering combinations, 1234 is the same as 2314 or 4123 etc. Actually... 4 numbers to be arranged with ten different possibilities each will give a total of 10^4 combinations, or 10,000.
There are 60C5 = 60!/(5!(60-5)!) = 60 × 59 × 58 × 57 × 56 / (5 × 4 × 3 × 2 × 1) = 5,461,512 There are 5,461,512 possible combinations of 5 numbers from 1-60.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
6 ways: 931,913,139,193,391,319
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
If you are able to repeat numbers you would take 6 * 6 * 6 = 216 combinations. If you are not able to repeat numbers you would take 6 * 5 * 4 = 120 combinations.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
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.
9
There are 1140 5 digit combinations from 1 to 20. 20 combination 3 computes that.
I believe the factors/digits would have to be either 1, 1, 2, 3 (12 combinations) or 1, 1, 1, 6 (4 combinations).
There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.