The questions are equivalent.
Your proposed answer to the first question is wrong.
Sanity Check
Since there are three choices for each of the five balls, there are $3^5 = 243$ ways to distribute five balls to three different boxes without restriction. Hence, the number of ways to distribute the balls so that there is at least one ball in each box must be at most $243$. However, $300 > 243$, so your proposed answer must be incorrect.
Five balls of different colours are to be placed in three boxes of different sizes. In how many ways can we place the balls so that no box remains empty?
Method 1: We use the Inclusion-Exclusion Principle.
There are three ways to choose a box in which to place each of the five balls. Hence, there are $3^5$ ways to distribute the balls without restriction.
From these, we must subtract those distributions in which fewer than three boxes receive a ball.
There are three ways to exclude one of the boxes from receiving a ball and $2^5$ ways to distribute the remaining balls to the remaining two boxes.
However, if we subtract $\binom{3}{1}2^5$ from $3^5$, we will have subtracted too much since we will have subtracted the three distributions in which all the balls are placed in one box twice, once for each way of designating one of the other two boxes as the empty box. Since we only want to subtract these cases once, we must add them back.
Thus, by the Inclusion-Exclusion Principle, the number of ways we can distribute five balls of different colours to three boxes of different sizes if each box must contain at least one ball is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 150$$
Method 2: We correct your attempt.
Three balls are placed in one box and one ball is placed in each of the other two boxes: There are three ways to choose which ball receives three balls and $\binom{5}{3}$ ways to select which three of the five balls are placed in that box. There are $2!$ ways to distribute the remaining two balls to the remaining two boxes. Hence, there are
$$\binom{3}{1}\binom{5}{3}2!$$
such distributions.
A single ball is placed in one box and two balls are placed in each of the other two boxes: There are three ways to select the box which receives a single ball and five ways to choose which ball is placed in that box. There are $\binom{4}{2}$ ways to choose which two of the remaining four balls are placed in the smaller of the remaining two boxes. The other two balls must be placed in the remaining box. There are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
such distributions.
Total. Since these cases are mutually exclusive and exhaustive, there are
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 60 + 90 = 150$$
ways to distribute five balls of different colours to three boxes of different sizes if each box must receive at least one ball.
Why is your approach incorrect?
Suppose the balls are blue, green, red, orange, and yellow. Moreover, suppose that the blue, green, and red balls are placed in the large box, the orange ball is placed in the medium box, and the yellow ball is placed in the small box. You count this case once when you place the orange ball in the medium box, then place the yellow ball in the small box, and the remaining balls in the large box, but you also count it when you place the yellow ball in the small box, then place the orange ball in the medium box, and the remaining balls in the large box.
What matters is which ball is placed in which box, not the order in which the balls are selected. You counted every case twice, once for each way you could put single balls in each of two boxes in the $(1, 1, 3)$ distributions and once for each way you could pairs of balls in each of two boxes in the $(1, 2, 2)$ distributions.
