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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.

Q: If you have 5 letters how many combinations can you make?

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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

Related questions

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