# Poker Probability with hands of 6 cards

I am trying to answer Question 1(d). A valid hand would be Ace hearts, 10 hearts, 2 clubs, 7 clubs, Ace spades, J spades.

My attempts have given me 2 different answers that I am unsure of.

What I did was I had to choose 3 suits from 4, then choose 2 cards from each of those suits. When choosing the 2 cards from each suit, do I have to do it 3 times? Would it be 4C3 x (13C2)^3 or 4C3 x 13C2

Thank you.

• Yes you have to do it three times, once for each of the three suits Mar 23, 2020 at 23:42

You are counting ways to select: two from thirteen kinds for each of three from four suits.$$({^{13}\mathrm C_2})^3\cdot{^4\mathrm C_3}$$
Note: $${^{13}\mathrm C_2}\cdot{^4\mathrm C_3}$$ would count ways to select the same two from thirteen kinds in three from four suits.
• @fatimahfatcakes Since there must be a suit with at least two kinds in the hand, we should assume that is asking for a suit with exactly two kinds . Then it becomes as easy as PIE (ie, use the Principle of Inclusion and Exclusion).$$\def\C#1#2{{^{#1}\mathrm C_{#2}}}\C 41\C{13}2\C{39}4-\C 42(\C{13}{2})^2\,\C{26}2+\C43(\C{13}{2})^3$$ Mar 24, 2020 at 0:33