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.
5! 5 * 4 * 3 * 2 *1 = 120 arrangements you take the number of letters in the words and make it a factorial.
Just 1.
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.
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."
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
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