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I'm trying to do this problem from "The Probability Tutoring Book" by Carol Ash.

If the letters in ILLINOIS are arranged at random, find the probability that the permutation begins or ends with L.

My attempt:

We'll first find the total number of permutations. Total Permutations: $\frac {8!}{(2!)(3!)}$

My reasoning for that total is we can arrange the eight letters in ILLINOIS $8!$ times. Since there are two L's and three I's, any arrangement where we just swap the I's with each other or the L's with each other, we have the same string(i.e. I$L_1L_2$INOIS is the same as I$L_2L_1$INOIS.)

Now I'll go about using the inclusion and exclusion principle to find the probability that our string begins with L or ends with L.

Total strings that begin with L: $\frac {7!}{(2!)(3!)}$

Where I use the same reasoning as above, but this time I fix the first letter to be "L" and can permute the other letters $7!$ ways

I use the same reasoning for the total strings that end with L : $\frac {7!}{(2!)(3!)}$

Now I take care of double counting, the total number of strings with both L as the first and last letter is $\frac {6!}{(2!)(3!)}$. Again, I applied the same reasoning as above.

Putting everything together,

$$P(\text {begins or ends with L}) = \left(\frac {1}{\frac {8!}{(2!)(3!)}}\right) \times\left(\frac {7!}{(2!)(3!)} + \frac {7!}{(2!)(3!)} - \frac {6!}{(2!)(3!)}\right) = \frac {13}{56}$$

However, according to the answers, the correct answer is $\frac {26}{56}$, so I'm off by a factor or 2.

A possible reasoning could be that I$L_1L_2$INOIS is NOT the same as I$L_2L_1$INOIS. Why would this be?

Attempting this again on my differently,

$P(\text{first letter is L}) = P(\text{last letter is L}) = \frac {2}{8}$ since there are 2 L's of 8 letters

$P(\text{first letter is L} \cap \text {last letter is L}) = \frac{2}{8}\frac{1}{7}$ since there are 2 L's for the first letter and then they'll be 1 L for the last letter.

Putting everything together:

$P(\text{first letter is L} \cup \text{last letter is L}) = \frac{2}{8} + \frac {2}{8} - \frac{2}{8}\frac{1}{7} = \frac {26}{56}$ which concurs with the answer in the book.

But I'm still struggling to figure out why my original answer was wrong. Did I undercount?

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  • $\begingroup$ Wait but L1_ _ _ _ _ _ L2 is the same as L2_ _ _ _ _ _ L1, so we should divide by two... no? $\endgroup$ Commented Jun 21, 2020 at 18:14
  • $\begingroup$ Oh! Are you saying that L_1 _ _ _ _ _ _ _ and L_2 _ _ _ _ _ _ _ each have arrangements (7! / 3!) because I specifically pick one of them to be at the start $\endgroup$ Commented Jun 21, 2020 at 18:17
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    $\begingroup$ An easier method would be to use the probability that it doesn't begin or end in an $L$. Namely find the answer as$$1-\frac{\binom{6}{2}\cdot6!/3!}{8!/(2!\cdot3!)}$$ $\endgroup$ Commented Jun 21, 2020 at 18:19
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    $\begingroup$ The number of ways to rearrange the $7$ letters $\{I,I,I,L,N,O,S\}$ is $7!/3!$. This equals the number of rearrangements starting (or ending) in $L$ because we just add the remaining $L$ to any of these arrangements. Although adding these double counts the cases with $L$ at the start and end. The number of ways to rearrange the $6$ letters $\{I,I,I,N,O,S\}$ is $6!/3!$. This equals the number of rearrangements starting and ending in $L$ because we just add the two $L$s to the start and end of any of these arrangements. $\endgroup$ Commented Jun 21, 2020 at 18:33
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    $\begingroup$ Yes. I don't see where this arrangement comes from though. $\endgroup$ Commented Jun 21, 2020 at 18:44

1 Answer 1

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You made the error when you calculate the total strings that start with $L$.

As you said, we can fix one $L$ in the front of the string. For example, $LILINOIS$. However, you calculated $\frac{7!}{2!3!}$ when it should be just $\frac{7!}{3!}$. This is because out of the seven remaining letters, all are distinct except for the $3$ I's. You would not divide by $2$ for the $2$ L's, since you've already fixed one in place. You can basically think that the first $L$ isn't even there, since you've already fixed it into position. No matter what permutation, it will always stay there. Same thing for the ending letter $L$ and the case of both starting and ending with $L$.

Everything else is correct. As expected, when calculated you get $\frac{26}{56}$ (or simplified, $\frac{13}{28}$).

Hope that helped,

-FruDe

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