Counting and Symmetry The digits $1,2,3,4$ and $5$ can be arranged to form many different $5$-digit positive integers with five distinct digits. In how many such integers is the digit $1$ to the left of the digit $2$? (The digits 1 and 2 do not have to be next to each other.)
Case 1: 2 is in the second spot
We get $3!=6$ ways.
Case 2: 2 is in the third spot
We get $4\cdot2=8$ ways.
Case 3: 2 is in the 4th spot
We get $3\cdot4=12$ ways.
Case 4: 2 is in the last spot.
We get $4!=24$ ways.
Thus, the probability should be $\frac{6+8+12+24}{5!}=\frac5{12}.$
Is this correct?

A different way is $\binom{5}{2}$ ways to pick 2 slots for 1 and 2, where 1 is always the left one, and $3!$ ways to arrange the rest of the numbers in their own 3 slots. So total number of cases is
$$
\binom{5}{2} \cdot  3! = 10 \cdot 6 = 60.
$$
This also changes the probability:
$$
\frac{60}{5!} = \frac{60}{120} = \frac12 = 50 \%.
$$

Now that you see a tedious way to find it's $50\%$, note there is an obvious symmetry. In the set of all $5!$ permutations, the number of cases where $1$ precedes $2$ is by symmetry the same as number of cases where $2$ precedes $1$, so the result must be $1/2$.
