# In how many different ways can you pick four cards so that your cards have exactly two suits?

Firstly, we pick which 2 suits we work with: 4 C 2

Secondly, there are three scenarios to consider:

Scenario 1: We pick 1 card from one suit and 3 from the other: 13 C 1 * 13 C 3

Scenario 2: We pick 2 cards from one suit and 2 from the other: 13 C 2 * 13 C 2

Scenario 3: We pick 3 cards from one suit and 3 from the other: $$13 C 3 * 13 C 1$$

(For instance, if we picked the suits to be diamonds and hearts, then from scenario one we pick 1 from diamonds and 3 from hearts and in scenario 3 we pick 3 from diamonds and 1 from hearts).

So the final answer is $$4 C 2* (( 13 C 1 * 13 C 3 )+( 13 C 2 * 13 C 2 )+( 13 C 3 * 13 C 1 ))$$

Is this the right way of approaching this problem?

• Looks reasonable to me Sep 10, 2020 at 22:09
• Yes, that how you do it and that is the correct answer (assuming you don't care about the order of the cards within the hand). But it's a small enough number I'd like so see some actual values and a final number. That is $_4C_2 = 6$ and $_{13}C_1 = 13$ and the whole thing is. Sep 10, 2020 at 23:33

$$\def\C#1#2{\mathop{^{#1}\mathrm C_{#2}}}$$You are counting ways to select 2 from 4 suits, and either 2 from both, or 3 from one and 1 from the other.

$$\C42\,(\C{13}2\C{13}2+2\C{13}3\C{13}1)$$

That is correct.

You can also count ways to select 2 from 4 suits, and 4 from the 26 cards in those suits except where all 4 are from either single suit.

$$\C42(\C{26}4-2\C{13}4)$$

(Re: Principle of Inclusion and Exclusion.)

Either way, it is $$81\,120$$