Consider a random sample of 11 letters chosen from the alphabet with replacement. What is the probability that the letters can be arranged to spell ‘mississippi’.

For this question, the number of ways to choose 11 letters would be (26)^(11). To spell 'mississippi',we need: 1M, 4I. 4S and 2P.

Hence, there are 11! ways of getting these specific letters. Hence, the probability of spelling 'mississippi' would be:

[11!] divided by [(26)^(11)].

Is this correct? Any alternative way of thinking about this?

  • $\begingroup$ Hence, there are 11! ways of getting these specific letters. <-- Why? No. $\endgroup$ – zipirovich May 26 '20 at 1:15

Your denominator of $26^{11}$ is correct, but your numerator should use the multinomial coefficient $\displaystyle \binom{11}{1,4,4,2} = \frac{11!}{1!4!4!2!}$.

Why? Well, the multinomial coefficient $\displaystyle \binom{n}{n_1, n_2, ..., n_k}$ tells us how many ways there are to put $n$ objects into $k$ groups of sizes $n_1, n_2, ..., n_k$, where the sizes add up to $n$. For example, we might have $10$ different objects and we want to figure out how many ways there are to distribute them into $3$ bins of sizes $3, 5, 2$. Then the formula tells us there are $\displaystyle \frac{10!}{3!5!2!} = 2,520$ ways.

With the word "MISSISSIPPI", you can think of there being $11$ spots, and we have to assign each of these spots to a letter -- one of M, I, S or P. There is 1 M, 4 I's, 4 S's, and 2 P's. So, we can think of this question as asking how many ways are there to take $11$ objects and putting them into four groups of sizes 1, 4, 4 and 2. This is simply the multinomial coefficient $\displaystyle \binom{11}{1,4,4,2} = \frac{11!}{1!4!4!2!} = 34,650$.

Edit: Alternatively, another way to think about it is, as another answer pointed out, as a permutation with repetitions. You have to divide by those factorials so as to remove repeats.


Your denominator is correct, but your numerator of "$11!$" isn't.

This — the MISSISSIPPI thing — is (imho) the most classis example of permutations with repetition. You have several objects, say $n$ of them, and you need to find the number of ways to permute them. If all $n$ object are different, then the answer is indeed $n!$. But notice that here you have groups of identical objects, such as four I's. Is you switch any two of them with each other, you won't get a different permutation. To account for that, you can use the permutations with repetitions formula.

  • $\begingroup$ I thought of doing that initially; i.e 11! divided by 4!*4!*2!*1! and then dividing that by 26^11, but I thought that even the identical letters such as the i's are going to be considered to be different. Since you want 4Is in total, each one is separate from the other. $\endgroup$ – jeff123 May 26 '20 at 1:31
  • $\begingroup$ @jeff123: The point is that they are indistinguishable from each other, so we can't treat them as different. $\endgroup$ – zipirovich May 26 '20 at 1:33
  • $\begingroup$ Fair enough. In terms of different arrangements of the word Mississippi, I 100% agree that you have to divide 11! by 4!*4!*2!*1!, as you said, in that case we are double counting. However, when it comes to selecting them, what I am confused about is that for example, if we want to spell the word "matt". Lets call the first t - t1 and the second t - t2. For say that out of all the arrangements, one is: A M T1 T2 and we also have A M T2 T1. $\endgroup$ – jeff123 May 26 '20 at 1:47
  • $\begingroup$ @jeff123: Here's one of them: amtt. Can you tell which one is it: the first or the second? $\endgroup$ – zipirovich May 26 '20 at 1:49
  • $\begingroup$ It could be either one of them $\endgroup$ – jeff123 May 26 '20 at 1:50

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.