There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls).
The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 \choose 6$ ways the first type of good event can occur.
The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 \choose 10$ ways for the second type of good event to occur.
Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).
Does this answer your question?