Answer:
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.
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.
