The only ways to get 24 is if we have \(1, 2, 3, 4\), \(1,1,3,8\), or \(1,1,4,6\)
For the first case, because all the digits are distinct, there are \(4! = 24 \) ways to factor.
For the second case, we have to account for overcounting by dividing by \(2!\)(the amount of ways to order 2 1s), giving us: \({4 ! \over 2!} = 12\)
LIkewise, the third case has 12, by the same logic.
Now, can you take it from here?