# Counting exactly 5 spades from 6 cards

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 left hand side is the probability the six cards drawn are made up of $5$ spades and $6-5=1$ from the rest of the deck. The right hand side seems to be the probability the thirteen spades are made up of $5$ cards of the six drawn and $13-5=8$ from the non-drawn cards. Feb 20 at 21:43

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: $$\frac{\binom{6}{5}\binom{52-6}{13-5}}{\binom{52}{13}}$$