We first make an assumption that you are picking 5 letters from 26. If this is not the case, the problem is different. This is covered at the end of the answer.
With repetition allowed (letters can be re-used), and order is not important since it is a combination (not a permutation), it is (26+5-1)!/(5!)(26-1)! This is a really big number!
If it were a permutation with repetition, this is the same as the number of ways to pick 5 things from 26 with replacement. There are 26 choices for each letter, so you
have 26x26x26x26x26 permutations.
No repetition, order is important, it is 26! / (26-5)! = 26 x 25 x 24 x 23 x 22 = 7,893,600
No repetition, order doesn't matter (ABCDE is the same as EDCBA is the same as CEDAB, etc), it is 26! / (5!(26-5)!) = (26 x 25 x 24 x 23 x 22)/(5 x 4 x 3 x 2 x 1) = 65,780
What if you just have 5 letter already picked. Meaning you are not worried about picking the 5 from the alphabet, you already have 5 fixed letter. Some may be the same letter or not.
If you allow for repetition of the letters, you have 5x5x5x5x5 permutations.
For combinations, it would be 11!/5!4!=7983360
MOST LIKELY, you just have 5 distinct letters and you want to arrange them where order does matter and you cannnot repeat.
In this case you have 5 choices for the first letter, 4 for the second, 3 for the third etc. So you have 5! or 120 choices.
Of course if you have 5 things and the order does not matter, there is only 1 way to arrange them. For example. ABCDE is the same as ABEDC.
So in dealing with combinations and permuations, the language must be quite exact. A small difference changes the entire problem.
See related MathsIsFun link for more information.
6P4 = 6!/(6-4)! = 6 * 5 * 4 * 3 = 360 four letter permutations from 6 different letters.6C4 = 6!/[4!∙(6-4)!] = 15 four letter combinationsfrom 6 different letters.
There are 120 ways to order a 5 letter word, however, since there are two "e" letters in the word "green", the number of combinations is only 60.
If digits can be reused, then 5x5x5x5x5 = 3125. If digits cannot be reused, then 5x4x3x2x1 = 120.
Just 1.
5! 5 * 4 * 3 * 2 *1 = 120 arrangements you take the number of letters in the words and make it a factorial.
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
Only one.
one hundred and twenty
20 different combinations of silverware
3 x 5 = 15
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.
5