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.
The word "space" consists of 5 letters, all of which are unique. To find the number of combinations (or permutations) of these letters, you can calculate 5! (5 factorial), which equals 120. Therefore, there are 120 different ways to arrange the letters in the word "space."
5C3 = 10
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.
20
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.
To find the number of four-letter words that can be created from the letters in "olvge," we first note that there are 5 unique letters. We can choose 4 letters from these 5 in ( \binom{5}{4} = 5 ) ways. Each selection of 4 letters can be arranged in ( 4! = 24 ) different ways. Therefore, the total number of four-letter combinations is ( 5 \times 24 = 120 ).
5C3 = 10
one hundred and twenty
Only one.
20 different combinations of silverware
3 x 5 = 15
5