There are 7C4 = 7!/(4!*3!) = 7*6*5/(3*2*1) = 35 combinations.
There are 1285C4 = 1285*1284*1283*1282/(4*3*2*1) = 113076300485 combinations.
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.
To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.
There are 11,238,513 of them.
4*3*2*1 = 24 different combinations.
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
There are 7C4 = 7!/(4!*3!) = 7*6*5/(3*2*1) = 35 combinations.
There are 1285C4 = 1285*1284*1283*1282/(4*3*2*1) = 113076300485 combinations.
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.
10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.
it is hard to say there are lot of combinations belive or not * * * * * If the previous answerer thinks 15 is a lot then true. There are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations. Not so hard to say!
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
4 of them. In a combination the order of the numbers does not matter.
One, in each case without leading 0s. 3 and 4 with leading 0s.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
Assuming a number can appear only once, there are 47C5 = 47*46*45*44*43/(5*4*3*2*1) = 1,533,939 combinations.