Easily Even

This problem can be basically rephrased into looking for all the ways to put the $5$ digits into $4$ places, so the that last place can only be a $2$ or a $4$ . Therefore, we get:

$1^{\circ}$ For the last digit we have $2$ possible options (either $2$ or $4$ ).

$2^{\circ}$ For the second digit (from the right) we have $4$ possible options (since repetition is not allowed and we already used one digit).

$3^{\circ}$ For the third digit (from the right) we have $3$ possible options (same as above, we already used two digits).

$4^{\circ}$ For the fourth and final digit (from the right) we have $2$ possible options (since we already used $3$ out of $5$ digits).

Thus, to sum up, the total number of combinations (and number of such integers) is $2\cdot 4\cdot 3\cdot 2 = \boxed{48}$ .

No of numbers possibly formed is 5! = 120. Probability last digit is even is 2/5. So no if such numbers is 2/5*120=48.

Since we know the last digit must be 2 or 4 (two options), we multiply the number of possible arrangements of the four other digits (4!) by 2. => 4!*2 = 48

$4\times3\times2\times 2=\boxed{48}$