# How do I find the probability of specified events from a permutation of the 26 english letters?

I found a similar problem here, but I don't really understand the explanation to their solution and can't apply it.

Question:

What is the probability of the following even when we randomly select a permutation of the 26 lowercase letters of the English alphabet?

a.) a is the first and z is the last in the permutation

The first and last letters are specified so the sample is 26-2=24 and the event size is the total letters in the alphabet (lowercase) which is 26. So $P(E)=\frac{24!}{26!}$?

b.) a and z are separated by at least 22 letters in the permutation

This is where I don't understand the Berkeley explanation of a similar scenario. I think there are 3 possible cases of separation: a and z are separated by exactly 22,23,or 24 letters? The rest Im fairly confused on working out.

Did I do part a correctly? And can someone help me with part b? Thanks!

• Yes, (a) is fine. More simply, we could say that the probability the first letter is an $a$ is $\frac{1}{26}$. Given this, the probability the last letter is a $a$ is $\frac{1}{25}$, for a probability of $\frac{1}{26}\cdot \frac{1}{25}$. – André Nicolas Apr 2 '14 at 1:26
• @AndréNicolas I'm glad you said that. For these types of problems it seems simpler to just think, "If I placed the letter a, I had 26 possible places for it to go and then if I place z, there are 25 remaining places for it to go." – joshmcode Dec 2 '15 at 2:02

We count the number of ways in which $a$ and $z$ are separated by $22$ or more, that is, by $22$, $23$, or $24$. We might as well count the patterns in which $a$ is first, and double.
Separation 24: There are $2$ patterns.
Separation 23: If $a$ is first, it can be in any of the first two positions. Then the position of $z$ is determined. There are $2(2)$ patterns.
Separation 22: The same reasoning shows there are $6$ patterns.
The total number of patterns is $12$. Any particular placement of $a$ and $z$ has probability $\frac{1}{26}\cdot \frac{1}{25}$, so the required probability is $12\cdot\frac{1}{26}\cdot\frac{1}{25}$.