What is the probability of the event that z precedes both a and b when we randomly select a permutation of the 26 lowercase letters of the English alphabet?
Currently, my thoughts are:
P(25,25)x24 There are p(25,25) ways to get certain letters of the alphabet
24 ways to place z, it can't be in the last 2 spots since it has to precede a
and b.
Best Answer
There are $6$ permutations of our set of $3$ letters. In $2$ of them, z precedes both a and b, so the probability is $\dfrac{2}{6}$.
The locations of the other letters is irrelevant. If this is not obvious, note that given any specific locations for the other $23$ letters, all of the $3!$ orders of our target letters are equally likely, and $2$ of these orders, z, a, b and z, b, a satisfy our condition.
Remark: One can also do it the hard(er) way. There are $26!$ permutations of the letters, all equally likely. Now we count the permutations in which $z$ precedes $a$ and $b$. There are $\dbinom{26}{3}$ ways to decide on the set of locations that will be occupied by the set of letters a, b, c. Once we have done that, there are $2$ ways to arrange a, b, and z in these locations so that z is first. And then there are $23!$ ways to arrange the remaining letters. So our probability is $$\frac{\binom{26}{3}(2)(23!)}{26!}.$$ This simplifies to $\dfrac{1}{3}$.