-2
-8
32C3 = 4960
100! / 97! = 100 * 99 * 98 = 970200
45
6
-2
-8
32C3 = 4960
Six combinations: 123, 132, 213, 231, 312, 321
i guess the answer is 54. A group of three digits can be selected from the 5 digits in 9 ways and each group can be arranged in 3! (i.e., 6) ways respectively and hence the answer is 6*9=54
100! / 97! = 100 * 99 * 98 = 970200
Only 1. In a combination, the order of the digits does not matter.
45
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
35,152,000 (assuming that 000 is a valid number, and that no letter combinations are disallowed for offensive connotations.) Also, no letters are disallowed because of possible confusion between letters and numbers eg 0 and O.
10,000.