How to calculate this probability more rigorously? I was working on a simple question that asked:
"Randomly select one of the $120$ permutations of the letters $a, b, c, d,$ and $e$. Find the probability that in the chosen permutation, the letter $a$ precedes the $b$ (the $a$ is to the left of the $b$)."
I know intuitively that the probability is $1/2$, by taking any arbitrary arrangement of letters, i.e., if $a$ is to the left of $b$, then reversing them will yield $b$ to the left of $a$ via a bijective mapping. Thus, half of all the possible arrangements of $a, b, c, d, e$ would give a to the left of $b$. However, I got stuck attempting to solve this using a more calculative approach.
If the sample space $n(S) = 120$, how can I find the number of positions/events that yield $a$ to the left of $b$ (which presumably would be $n(E)=60$, since the probability is $1/2$)? Thank you.
 A: Choose any two places from the five places . This can be done in $5 \choose 2$ ways . After choosing these two places , we always place $'a'$ on the left one and $'b'$ on the right one from the 2 chosen places . Also the number of possible permutations of the letters $c,d,e$ in the remaining 3 places will be $3!$ . Therefore , the total number of permutations in this way will be $${5\choose2}3!= 60$$ways.
We also know that $n(S)=120$ , therefore probability that in the chosen permutation , $a$ precedes $b$ is $$\frac{60}{120}=\frac12$$
A: Your given approach is actually preferred, and sufficiently rigorous if your construct the bijection, but there are many ways to solve this.
$(1)$ Let $\{p_1,p_2\}$ be the set of positions taken by $a$ and $b$ in the arrangement.  This can take  ${5 \choose 2} = 10$ values, and each being occupied uniquely by $a$ and $b$ due to the ordering constraint. The remaining $3$ positions can be occupied randomly by the remaining $3$ numbers, giving a total of ${5 \choose 2} * 3! = 60$ positions.
$(2)$ Let $P_{k}$ be the number of ways to arrange $\{a,b\}$ and another $k$ elements as a sequence of length $k+2$ such that $a$ precedes $b$. We have the following recurrence relations:
$$P_{0} = 1\\
P_{k} = (k+1)! \; + \; k * P_{k-1} 
$$
Ie. when we have no additional elements $(k=0)$, $\;\;a,b\;\;$ is the only valid sequence. When we have $k$ additional elements, we can split the sequences into those where $a$ is the first element (a total of $(k+1)!$ as all sequences with $a$ first are valid) and those which do not have $a$ first giving $k$ choices of the first element as $b$ can never be first, and the remainder $k-1$ length subsequence being counted by $P_{k-1}$.
We compute: $$P_0 = 1\\ P_{1} = 2! + 1*P_0 = 3 \\  P_{2} = 3! +  2 * P_{1} = 12\\
P_{3} = 4! + 3 * P_{2} = 60$$
As $P_{3}$ counts the number of sequences of $a,b$ and $3$ additional elements where $a$ precedes $b$, it is the number we are looking for.
