How many 3 number combinations can you make from 1 2 3 4 5 using each number once?

Answer 1

To calculate the number of 3-number combinations from the set {1, 2, 3, 4, 5} using each number once, we can use the permutation formula. Since order matters in a combination, we use the permutation formula nPr = n! / (n-r)! where n is the total number of items and r is the number of items chosen. In this case, n = 5 (total numbers) and r = 3 (numbers chosen), so the calculation would be 5P3 = 5! / (5-3)! = 5! / 2! = 5 x 4 x 3 = 60. Therefore, there are 60 different 3-number combinations that can be made from the set {1, 2, 3, 4, 5} using each number once.

Answer 2

123, 124, 125, 132, 134, 135, 142, 143, 145, 152, 153, 154

213, 214, 215, 231, 234, 235, 241, 243, 245, 251, 253, 254

312, 314, 315, 321, 324, 325, 341, 342, 345, 351, 352, 354

412, 413, 415, 421, 423, 425, 431, 432, 435, 451, 452, 453

512, 513, 514, 521, 523, 524, 531, 532, 534, 541, 542, 543

A total of sixty combinations.

You could also use the Permutation formula with no repetition to get the result, without having to list them all out. This is important once you start getting into higher numbers.

The formula is n! / (n-r)! where n is the total number of items, and r is the number of items you are taking for each permutation.

The exclamation (!) means factorial. You take n * (n-1) * (n-2) * (n-3) * . . . . * 1.

In this case n=5 and r=3

so 5! / (5-3)! = (5 * 4 * 3 * 2 * 1) / (2 * 1) = 120/2 = 60. I'm posting a link for more information.