Suppose the 5 letters are A, B, C, D and E.
The letter A can either be in the combination or not: 2 options for A.
With each of these options, B can either be in the combination or not: 2 options for B - making 2*2 options so far.
With each of the options so far, C can either be in the combination or not: 2 options for C - making 2*2*2 options so far.
and so on.
So for 5 letters there are 25 = 32 combinations. However, one of these is the combination that excludes each of the 5 letters - ie the null combination. Excluding the null combination gives the final answer of 31 combinations.
120 ABCDE , there are 120 ways these five letters can be rearranged.
5C3 = 10
56C5 = 3,819,816
one hundred and twenty
3 x 5 = 15
7,893,600 (seven million, 8 hundred ninety-three thousand, 600) combinations in English.
120 ABCDE , there are 120 ways these five letters can be rearranged.
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.
The first digit can be one of five integers (1 to 5) The second - fourth digits one of six integers (0 to 5). So: number of (valid) combinations is 5*6*6*6=1080 * * * * * This gives the number of permutations, not combinations. I am working on the latter.
20
5C3 = 10
5. One each with one of the 5 letters left out: abcd, abce, abde, acde and bcde.
Only one.
56C5 = 3,819,816
one hundred and twenty
20 different combinations of silverware
3 x 5 = 15