How many permutations of $\{1,2,3,4,5\}$ begin with an even number?

Question

I came across the following problem, and I am looking for some help,

How many permutations of $\{1,2,3,4,5\}$ begin with an even number?

My workings,

So in the set of numbers there are $2$ even numbers, $2$ and $4$, and we must fill $5$ positions with the beginning number being even,

We have $2$ options for the first position, $5$ options for the second position, $5$ options for the third position, $5$ options for the 4th position, and finally $5$ options for the last position, so therefore by the product rule

$$2\times5\times5\times5\times5 = 2 \times 5^4 = 2\times625 = 1250$$

Answer 1

You'd be correct if the question said you can reuse numbers. However, in a permutation, you cannot repeat numbers.

With this in mind, we have $\color{blue}2$ options for the first position, $\color{blue}4$ options for the second position, $\color{blue}3$ options for the third position, $\color{blue}2$ options for the 4th position, and only $\color{blue}1$ option for the last position, so therefore by the product rule we have $$2\times4\times3\times2\times1 = 48$$ permutations.

More succinctly, we have $\color{blue}2$ options for the first position, and we wish to permute the rest of the $\color{blue}4$ numbers available to us. Hence we have

$$2 \times 4! = 48$$ even-leading permutations.

Answer 2

For the first position you have two choices as you said. But we do not consider repetition in permutations. Hence the other digits can be permuted in $4!$ ways.
Hence the answer would be $2*4!=48$