# Inclusion-Exclusion combinatorics question

This is the first question I answer on the subject so I require assistance.

We are asked the following:

In how many different ways can I draw 6 cards out of a deck with 52 cards, such that i drew at least 1 of each shape (hearts, spades, diamonds, clubs).

can anyone give me a hint in the right direction?

I have an idea but it seems very difficult, perhaps there is a better way.

I'm thinking of saying $DD$= set of 4 different shapes + 2 diamonds, $DH$ = set of 4 different shapes + another heart + another diamond,$DS$ 4 of each + another diamond + another spade, etc...find the cardinality of all these sets, and invoke inclusion-exclusion. But that seems very very difficult.

• How about: give numbers $I=1,2,3,4$ to the shapes. Then let $E_i$ be the event that you lack card $i$ in your draw? – user99680 Dec 8 '13 at 22:40

## 1 Answer

This is my suggestion: label your suits by $1,2,3,4$ , then let $E_i$ be the event of drawing a hand of $6$ cards where suit $i$ is missing. Then you want to count the total number of hands minus $|E_1 \cup E_2 \cup E_3 \cup E_4|$

• that is brilliant. great advice. Much simpler than what I proposed. – Oria Gruber Dec 8 '13 at 22:44
• Glad to help, Oria. – user99680 Dec 8 '13 at 22:45
• @user99680 can you elaborate on the event $E_i$ ? Do you mean $4{39 \choose 6}$ ? – GinKin Dec 10 '13 at 14:29