# How many possible outcomes of getting 2 face values when drawing 6 cards from a deck of 52?

We draw six cards from a deck of 52 playing cards. How many possible outcomes of getting 2 face values?

I don't understand what it means when it says "how many possible outcomes of getting 2 face values." Face values are any value 1,2,3...,J,K,A, correct? So if you only get 2 face values is it asking how many possible outcomes of 2 3-of-a-kinds?

The answer is 13C2 [2 x 4C2 + (4C3)^2]. Could someone explain the thought process in getting to this solution?

• Face cards have a face on them. Queen, Jack, and King. Feb 11, 2016 at 18:15
• I'd have said the "face cards" were only $J,Q,K$.
– lulu
Feb 11, 2016 at 18:15
• I would have thought that, but that definition wouldn't make sense in the context of some of the other questions in the question set. I'm pretty positive it is intended to simply mean different numbers. Feb 11, 2016 at 18:17
• Given the $\binom{13}{2}$ in the intended answer, it seems that they are looking for solutions of $XXXXYY$ or $XXXYYY$ Feb 11, 2016 at 18:19
• @juleshk13 What are the other questions for which the definition of face cards as $J, Q, K$ wouldn't make sense? Feb 11, 2016 at 18:31

There are 13 values, so the there are $C(13,2)$ ways to select X and Y. Now, in the two cases XXXXYY and XXYYYY we only have to choose which 2 cards (YY or XX) are selected, because the other are all taken, this gives the $2 \cdot C(4,2)$.
For XXXYYY we have to select 3 out of 4 for both X and Y so $C(4,3) \cdot C(4,3)$, and this gives us the result $C(13,2) \cdot (2 \cdot C(4,2) + C(4,3)^2 )$.