4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Hint: A ten-letter word whose letters are in alphabetical order is completely determined by the number of each letter it contains. So if you let $n_A$ be the number of $A$s, $n_B$ be the number of $B$s, etc., you want the number of nonnegative integers solutions to the equation $n_A+n_B+\cdots+n_Z=10$. The solutions are in one-to-one correspondence with the words you want to count. $\endgroup$ – Steve Kass Sep 23 '18 at 17:03
  • $\begingroup$ I'm starting to feel like you're suggesting that this has become into some kind of stars and bars question instead of counting. Though, I'm not going to lie, I'm having trouble seeing the connection between this problem and stars and bars. If possible, could you go further into detail? $\endgroup$ – DevAllanPer Sep 23 '18 at 17:20
  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 23 '18 at 19:03
3
$\begingroup$

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 $\binom{35}{10}$.

$\endgroup$
  • $\begingroup$ That makes sense! I appreciate it! Out of curiosity, how would the problem change if repetition were not allowed? $\endgroup$ – DevAllanPer Sep 23 '18 at 17:23
  • 1
    $\begingroup$ Then it would be even easier. All you would need to do is choose the $10$ letters that will appear in the word, so $\binom{26}{10}$ options. $\endgroup$ – Mark Sep 23 '18 at 17:24
  • $\begingroup$ I apologize, I don't see how that would guarantee that the letters would remain in listed in alphabetical order $\endgroup$ – DevAllanPer Sep 23 '18 at 17:27
  • 1
    $\begingroup$ Once you choose which $10$ letters will appear you have only one way to order them (they must appear in alphabetical order) so you have the word. By the way, I just noticed your answer to part b) of the question was wrong. There the question was no repetitions and any order of letters. So there you needed to choose $10$ letters outside of $26$ and then choose their order which is $10!$ options. So the answer to b) is $\binom{26}{10}\times 10!$. $\endgroup$ – Mark Sep 23 '18 at 17:29
  • 1
    $\begingroup$ Oh, right. Sorry, I thought by $P(26,10)$ you meant $\binom{26}{10}$. Got confused. Yes, then it was correct. But yes, if you need no repetitions and alphabetical order then the number is $\binom{26}{10}$. You just choose the $10$ letters and can't choose the order in that case. $\endgroup$ – Mark Sep 23 '18 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.