There is 345, 354, 435, 453, 543, 534.
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.
There are 120 permutations and 5 combinations.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
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.
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.
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!
654321-100000= 554321 combinations
10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210