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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.

ClownInTheMoon
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  • How do you get that? If I just take the "no club" case there are $39$ non-clubs so there are $P(39,7)$ for that case already. – lulu Dec 02 '16 at 22:53
  • @lulu Well I was counting trying to the count the number of cards with no club and no red card. But the answer $P(39,7)+P(26,7)$ is also incorrect according to the book. – ClownInTheMoon Dec 02 '16 at 23:05
  • In 'no card is a club or no card is red', is that really an 'or', is that an 'and'? – Fimpellizzeri Dec 02 '16 at 23:05
  • @Fimpellizieri this is the question verbatim from the book. – ClownInTheMoon Dec 02 '16 at 23:06
  • @ClownInTheMoon Yes..because now you are double counting the cases in which you have both no club AND no red card (i.e. all spades). – lulu Dec 02 '16 at 23:07
  • @lulu Sorry, I should've clarified that this answer is also incorrect $P(39,7)+P(26,7)-P(13,7)$. I've tested a bunch of answers which I would think are "logical" but the answer given is much larger. – ClownInTheMoon Dec 02 '16 at 23:08
  • @lulu The answer given is $P(52,7)-P(13,7)-P(26,7)$. – ClownInTheMoon Dec 02 '16 at 23:10
  • The answer given looks wrong. I found the same answer you gave in the previous comment, Validated by inclusion-exclusion. – Fimpellizzeri Dec 02 '16 at 23:17
  • @Fimpellizieri but in your answer it looks like you are using combinations and not permutations. Order matters in this question. – ClownInTheMoon Dec 02 '16 at 23:18
  • Oh, so the order matters? Just multiply the result by $7!$. – Fimpellizzeri Dec 02 '16 at 23:20
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    The answer given suggests the interpretation of the condition on the 7 cards was "It's not the case that all the cards are clubs; furthermore, it's also not the case that all the cards are red". That's an odd way to interpret the sentence they wrote, but, shrugs… – Sridhar Ramesh Dec 02 '16 at 23:48

1 Answers1

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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

  1. $\binom{39}{7}$ ways to choose $7$ cards such that none of them are club.
  2. $\binom{26}{7}$ 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.

Fimpellizzeri
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