Given the word "MISSISSIPPI", consider each of the letters of the given word once. Each letter can differ from each other because of uppercase/lowercase and color ( for example green A is different from blue A ). You can choose 8 different colors. What is the probability that if you arrange these letters randomly, you get the word "MISSISSIPPI" again? ( it doesn't have to be uppercase and there's no restriction on colors, as long as the word is the same ). The result should be $\frac{1}{34650}$. This is how I tried to solve it, but I couldn't get the exact result.

MISSISSIPPI is composed of 4 distinct letters: M-I-S-P.

Since each letter can differ because of color and uppercase/lower case we have these amount of choices for the letters:

4 distinct letters * 2 character types ( uppercase/lowercase) * 8 different colors = 64 different letters

The probability of arranging the words and getting MISSISSIPPI again is: $P(A) = \frac{number\space of\space arrangements\space that\space contain\space the\space word\space mississippi}{total \space arrangements}$

I calculated the number of total arrangements like this:

We have a total of 64 different letters taken 11 at a time, because the word MISSISSIPPI is composed of 11 letters. Since repetitions are allowed and order matters, we have $64^{11}$ total arrangements of these 64 letters because we have to do 11 choices and we have 64 options for each choice.

Now let's calculate in how many arrangements we get the word "mississippi".

We still have to make 11 choices with repetitions and order.

First letter has to be an M, it can be uppercase/lowercase and can be of 8 different colors, so we have

1 letter * 2 character types (uppercase/lowercase) * 8 colors = 16 different way of getting an M

Same reasoning can be done with every letter of the word "missisippi", so we have 16 options for each choice every time and we have a total of 11 choices, so the number of arrangements with the word "missisippi" are $16^{11}$.

Let's calculate $P(A)$

$P(A) = \frac{16^{11}}{64^{11}} = \frac{2^{44}}{2^{66}} = \frac{1}{2^{22}} = \frac{1}{4194304} $

What did I do wrong?

  • 2
    $\begingroup$ I don't understand all the stuff about the colors and the upper and lower case. It seems to me that you are just asking "if you take a random permutation of the letters in the given word, what is the probability that it coincides with the original word?" If so, then the numerator is just $1\times 4\times 4\times 3\times 3\times 2\times 1\times 2\times 2\times 1\times 1$ and the denominator is $11!$. $\endgroup$ – lulu Sep 3 '20 at 13:55

As lulu, I also don't really understand the stuff about colours and upper/lowercase.

By the way, MISSISSIPPI contains 11 letters which can be permuted in $11!$ possible way. So $11!$ is the number of all possible cases.

The character $S$ has four recurrences, so at least $4!$ of all possible permutations leave MISSISSIPPI invariant, namely we permute $S$ is all possible ways leaving the other characters fixed!

Same discussion for the character $I$, so there are other $4!$ permutations which leave MISSISSIPPI invariant. None of these permutation moves $S$!

Finally, $P$ appears two times and then there are other two permutations leaving MISSISSIPPI invariant.

We now want to permute the characters $S$, $P$ and $I$ at the same time. Therefore by combine all the permutations we have found above. These are $4!4!2!$.

Therefore the probability of leaving MISSISSIPPI invariant is $\displaystyle\frac{4!4!2!}{11!}=\frac{1}{34650}$ as desired.

  • $\begingroup$ Thanks, very clear answer $\endgroup$ – Tom Avery Sep 3 '20 at 19:51

Uppercases and colors are just a smokescreen. Essentially, you choose the order of the letter and then do whatever you want to them (color them...), but the second stage doesn't matter.

So, all you are left with is the number of different words which can be built: $\tfrac{11!}{4!2!2!}=34650$. Only one of them is Mississipi and that's the desired answer: $\tfrac{1}{34650}$.


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