How many three-digit even numbers can be formed using the digits 1, 2, 3, 4 and 5, when repetition of digits is not allowed?

For the number to be even, the last(or unit) digit has to be a multiple of 2.

In this case, the multiples of 2 present are 2 and 4.

So, let's say the last(or unit) digit is 2.

In this case we are left with 4 digits (1,3,4 and 5) and we have to fill 2 vacant places (the hundredth place and tenth place).

Number of ways of doing this = $$^4C_2\times\ 2!=12$$ ways

Similarly we can form a 3-digit even number with last(or unit) digit as 4, in 12 ways.

So, total number of ways = 12+12 = 24 ways.