In the question, we are given that the LHS of the equation (in the link above) is equivalent to the probability of drawing exactly 5 spades when drawing 6 cards from a standard 52 card deck, and we have to argue why the RHS is equivalent.
I noticed that the RHS is just the LHS with the positions of the 13s and 6s swapped. I'm thinking that for the RHS, 6 choose 5 represents a 6 card hand with 5 spades in it already, and 46 choose 8 represents the rest of the deck once those 6 cards, 5 of which are spades, have been drawn.
But I'm not too sure why the denominator needs to be 52 choose 13 and why the RHS as a whole is equivalent to the LHS. Also, we are explicitly told not to simply calculate both sides, we have to argue in words why they are equivalent.
The LHS partitions the $6$ chosen cards into $5$ spades and $6-5$ non-spades: $$\frac{\binom{13}{5}\binom{52-13}{6-5}}{\binom{52}{6}}$$
The RHS partitions the $13$ spades into $5$ chosen and $13-5$ not chosen: $$\frac{\binom{6}{5}\binom{52-6}{13-5}}{\binom{52}{13}}$$
These are both examples of the hypergeometric distribution. See https://en.wikipedia.org/wiki/Hypergeometric_distribution#Combinatorial_identities for a more general argument.