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Here's a question from my probability textbook:

All the combinations of $n$ letters, $r$ together, repetitions permissible, are written down. Find the chance that a combination selected at random will contain no repetition. ($n \ge r$)

Here's what I did. Clearly the numerator is $\binom{n}{r}$, and the denominator is going to be the number of nonnegative integer solutions to$$\ell_1 + \ell_2 + \ldots + \ell_{n-1} + \ell_n = r.$$However, I'm not sure how to compute that. I've tried messing with stars and bars, but didn't get anywhere. Can anyone give me a hint on how to proceed? No complete solutions please.

EDIT: There seems to be some confusion in the comments which maybe I've inadvertently contributed to. The problem statement says:

All the combinations of $n$ letters, $r$ together, repetitions permissible, are written down.

I interpret this as follows. If I have the $4$ letters $a$, $b$, $c$, $d$, and we're letting $r = 3$, then the possible combinations are:$$aaa, bbb, ccc, ddd, aab, aac, aad, bba, bbc, bbd, cca, ccb, ccd, dda, ddb, ddc, abc, abd, acd, bcd,$$for a total of$$4 + 4(3) + 4 = 20$$combinations written down.

EDIT 2: Okay, I ended up getting $\binom{n + r - 1}{r}$ for the number of nonnegative integer solutions to$$\ell_1 + \ell_2 + \ldots + \ell_{n-1} + \ell_n = r,$$so the desired probability is$${{\binom{n}{r}}\over{\binom{n + r - 1}{r}}} = {{n!(n-1)!}\over{(n+r-1)!(n-r)!}}.$$Is there a way to simplify this expression any further?

Emperor Concerto
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    What goes wrong with Stars and Bars? – lulu Sep 02 '21 at 16:53
  • To a form a combination, you have $n$ choices for each of the $r$ positions in the combination. That makes $n^r$ possible combinations. The ones without repetitions form a total of $\binom nr$ combinations as you wrote. – nejimban Sep 02 '21 at 16:54
  • Thanks @nejimban. I don't think what you wrote for the numerator, $n^r$, is quite right since $aaabb$ is the same combination as $bbaaa$. – Emperor Concerto Sep 02 '21 at 17:00
  • @EmperorConcerto: sorry I may have not understood your problem correctly then. – nejimban Sep 02 '21 at 17:03
  • Your desire to compute a probability is consistent with counting $aaabb$ as distinct from $bbaaa$ (in the denominator) provided you are sampling letters independently and with uniform distribution. The word combination is somewhat at variance with that interpretation, which to me says you should clarify the probability distribution rather than hiding its details behind the phrase "at random". Perhaps the selection of letters is done without replacement, thus making repetitions impossible. – hardmath Sep 02 '21 at 17:14
  • The specific random processes at play here are important. In many problems we might have picked the first letter... then picked the second letter... and so on, each choice made independently which leads to $n^r$ equally likely possibilities. However... here we wrote all the unique possibilities (where presumably order doesn't matter) down and picked from that list. This is one of the rare occurrences where the outcomes counted by stars-and-bars are actually equally likely to have occurred. – JMoravitz Sep 02 '21 at 17:23
  • So... count the number of possibilities with stars and bars. You ask for us not to give a complete solution... so I will leave you to find that count and learn about stars and bars yourself unless you ask for further clarification... but once you have that value you do very simply just take your numerator of $\binom{n}{r}$ and divide by that result and you are done. – JMoravitz Sep 02 '21 at 17:24
  • As a reminder... be careful as to the variable names and where they appear in the result for stars and bars. Some use $n$ as the number of balls with $r$ as the number of bins while others use $n$ as the number of bins with $r$ as the number of balls. Make sure you understand which is which in the formula you look at and whether your $n$ in your problem should have been the balls or bins in order to make sure everything fits together correctly. – JMoravitz Sep 02 '21 at 17:27
  • Hi all, I came up with an answer but I couldn't simplify it any further, see my edits. Is anyone able to help? – Emperor Concerto Sep 02 '21 at 18:13
  • Be careful. Which are balls and which are bins ? – true blue anil Sep 02 '21 at 18:19
  • @trueblueanil $\binom{n+r-1}{r}=\binom{r+n-1}{n-1}$ is indeed the correct solution for stars-and-bars here. The warning was given already before the attempt and the warning after the attempt is unwarranted and might be misinterpreted as though the OP got it wrong here. The OP got it right. @ OP, why do you want to simplify it further? This looks simple enough to me. That is more than acceptable as a final answer (and really, I'd prefer it with the binomial coefficients rather than with the factorials) – JMoravitz Sep 02 '21 at 21:57

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HINT

With your $4$ letter example, you have only $3$ objects to place any which way

$\boxed{A}\boxed{B}\boxed{C}\boxed{D}$

So the ans will be $\dfrac{\binom{n}{r}}{\binom{r+n-1}{n-1}}$

true blue anil
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