There are 120 permutations and 5 combinations.
all numbers have to be different 720 360 (6*5*4*3*2*1), * * * * * Wrong. Combinations do not have to be different. Those are permutations. There are only 6C4 = 6*5/(2*1) = 15 combinations.
120 different combinations (5 x 4 x 3 x 2).
654321-100000= 554321 combinations
There are 2*2*5 = 20 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.
There are 120 permutations and 5 combinations.
They are: 2*2*3*5 = 60
all numbers have to be different 720 360 (6*5*4*3*2*1), * * * * * Wrong. Combinations do not have to be different. Those are permutations. There are only 6C4 = 6*5/(2*1) = 15 combinations.
just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436
Assuming a number can appear only once, there are 47C5 = 47*46*45*44*43/(5*4*3*2*1) = 1,533,939 combinations.
120 different combinations (5 x 4 x 3 x 2).
654321-100000= 554321 combinations
There are 2*2*5 = 20 combinations.
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
To calculate the number of 4-number combinations using the numbers 1, 2, 3, 4, and 5 without repetition, you can use the formula for permutations. Since order matters in a combination, you would use the formula for permutations, which is nPr = n! / (n - r)!. In this case, you would have 5 choices for the first number, 4 choices for the second number, 3 choices for the third number, and 2 choices for the fourth number. Therefore, the total number of 4-number combinations would be 5P4 = 5! / (5-4)! = 5 x 4 x 3 x 2 = 120 combinations.