# [Math] How many $10$-letter words can be formed using the $26$ letters of the alphabet if repetition is allowed, but letters are listed alphabetically

combinatoricspermutations

How many $$10$$-letter words can be formed using the $$26$$ letters of the alphabet if:

a) Repetition is allowed

b) Repetition is not allowed

c) Repetition is allowed, but letters are listed alphabetically

I know that:

a) is $$26^{10}$$

b) is $$P(26, 10)$$

But I'm not sure where to start with c. Initially, I thought I could subtract $$10$$ from $$26$$ since you couldn't possibly form a word alphabetically if you didn't have enough letters but since repetition is allowed you could technically have the word 'ZZZZZZZZZZ'. So that rules that thought out. Any ideas as to how I should attack this problem?

Let's look at a different question first: how many ways are there to put $$25$$ bars between $$10$$ stars? (when between every two stars there can be as many bars as I wish). You need to take a line of $$35$$ places, choose the $$10$$ places where you want the stars to be, in the rest of the places there will be bars. So you just need to choose $$10$$ places from $$35$$ which is $$\binom{35}{10}$$ options.
Now, why is it equivalent to your problem? Because to create a word you need to choose how many times the letter $$a$$ will appear (zero times is also an option), how many times $$b$$ will appear and so on. Once you choose that for every letter you will get the word because you are not allowed to choose the order of the letters. So the number of times $$a$$ will appear is the number of stars before the first bar, the number of times $$b$$ will appear is the number of stars between the first and second bars and so on. The letter $$z$$ is letter number $$26$$ so it is the number of stars after the $$25$$th bar. So the number of words is $$\binom{35}{10}$$.