I am reading the following problem:
The digits $1$,$2$,$3$,$4$,$5$ are randomly arranged to form a three digit number (digits are not repeated). What is the probability that the number is even and greater than $500$?
My solution:
There are $5\cdot 4 \cdot 3$ ways to form the number in total.
In order to be greater than $500$ the first digit must be $5$ and in order to be even the last digit must be either $2$ or $4$. So there is $1$ way to select the first digit, $2$ ways to select the last and $2$ ways to select the middle one (either $3$ or $4$) i.e. $2\cdot 2\cdot 1$ = $4$ ways
Therefore the probability is: $\frac{4}{60}=\frac{1}{15}$
But the solution mentions $\frac{1}{10}$.
What am I doing wrong here?
J Moravitz' comment addresses the flaw.
The easiest alternative approach is that there are $D = 5!$ numbers possible, of which $4!$ of the numbers start with the digit $5.$ By symmetry, there are $N = \frac{4!}{2}$ such numbers that are even. So the desired probability is $\frac{N}{D}.$
Addendum
Per OP's request, I will combine his analysis with the analysis inherent in the comments of J Moravitz to provide an alternative direct approach to the problem.
As in the start of my answer, the computation will be
$$\frac{N}{D}, ~~\text{where}~~ D = 5!.$$
The leftmost digit must be $5$. The rightmost digit must be either $2$ or $4$. I will let $N_1$ denote the number of possibilities where the rightmost digit is $2$. Then, by symmetry, $N = 2 \times N_1$.
With the leftmost and rightmost digits being $5$ and $2$ respectively, the other 3 digits, $1,3,4$ can be permuted in $3!$ ways to fill in the 3 middle digits.
Therefore, $N_1 = 3!$ and $N = 2 \times 3!$.