How would one answer this simple permutations question? You deal 7 cards off of a 52-card deck and line them up in a row. How many
possible lineups are there in which no card is a club or no card is red?
My answer was $P(13,7)$ which is incorrect according to my book.
 A: The tool to use here is the Inclusion-Exclusion Principle.
There are $13$ club cards and $26$ red cards in a standard deck, so there are $39$ non-club cards and $26$ non-red cards. Hence, there are


*

*$\binom{39}{7}$ ways to choose $7$ cards such that none of them are club.

*$\binom{26}{7}$ ways to choose $7$ cards such that none of them are red.


Let $A$ be the set of $(1)$ and $B$ be the set of $(2)$. We wish to find $|A\cup B|$. By the inclusion-exclusion principle:
$$|A\cup B|=|A|+|B|-|A\cap B|$$
We know $|A|=\binom{39}{7}$ and $|B|=\binom{26}{7}$, so we need only calculate $|A\cap B|$.
Elements in $A\cap B$ are sets of $7$ cards such that each of those cards are simultaneously neither club nor red. In other words, elements in $A\cap B$ are sets of $7$ spades cards. There are $13$ spades cards, so
$$|A\cap B|=\binom{13}{7}$$
and we have our answer.
EDIT: If order matters, simply multiply the result by $7!$. Since each card is unique, each set of $7$ cards can appear in $7!$ distinct orders.
