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A single card is drawn from a standard 52-deck of cards with four suits: hearts, clubs, diamonds, and spades; there are 13 cards per suit. If each suit has three face cards, how many ways could the drawn card be either a club of any kind or anything else besides a face card?

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Hint: How many non-face cards are there? How many clubs are there? How any cards are both non-face cards and clubs? –  JavaMan Apr 19 '11 at 1:36
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this question sounds like a homework problem and if this is the case you may want to consider using the homework tag for this question. Also, when asking a question like this, its a good policy to outline your thoughts about the problem so that answerers can see where you got stuck and can thus provide better responses. –  WWright Apr 19 '11 at 1:36
    
I am assuming that n=52 and r=30, order doesn't matter so it's a combination???? –  user9762 Apr 19 '11 at 2:32
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Paula, you might also want to delete the other (~identical) copy of this question. –  The Chaz 2.0 Apr 19 '11 at 2:38
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2 Answers

The total number of possible outcomes is $52$, because that's the number of cards: you have one outcome per possible cards.

How many outcomes are "good" (that is, are "either a club of any kind or anything else besides a face card")?

Well, there's one "good outcome" per club; that's 13 clubs.

Of the remaining 39 cards, how many are also "good outcomes"?

Out of the 13 hearts there are, ten hearts (all hearts except the three face cards) count as "good outcomes." So in addition to the 13 clubs, 10 hearts are also "good."

How many spades are "good"? How many diamonds?

Now just add them all up.

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I don't think she is worried about probability here, just the "number of good outcomes". –  JavaMan Apr 19 '11 at 3:04
    
I feel like a total idiot. I made this problem much more difficult than what is there. Thanks DJC. I appreciate it. –  user9762 Apr 19 '11 at 3:05
    
@DJC: Oops. Quite right. –  Arturo Magidin Apr 19 '11 at 3:12
    
@Paula: No need to feel down. Talking about math with other people is useful! –  JavaMan Apr 19 '11 at 3:15
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I have a feeling your teacher/professor intended for you to learn the following (we first introduce some notation):

For two sets $A, B$ in a common universe $U$, define their union as

$$ A \cup B = \{x \in U : x \in A \text { or } x \in B\}. $$

Define their intersection as:

$$ A \cap B = \{x \in U : x \in A \text{ and } x \in B\}. $$

It is important not to get too wrapped in the English here. Being in $A \cup B$ simply means being in one of the two sets (or possibly both). Being in $A \cap B$ simply means being in set $A$ and being in set $B$ at the same time.

Finally, a finite set $A$ with $k$ elements ($k$ things in the set) has cardinality $k$, and this is written as $|A| = k$, or $\#A = k$ or even sometimes, $n(A) = k$.

Therefore:

$$ \{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\}, $$

while

$$ \{1 ,3 , 5 \} \cap \{ 1 , 2, 3 \} = \{ 1, 3\}. $$

Also, $$ | \{ 1,3,5\}| = 3, $$

while

$$ | \{ 1, 3\} | = 2. $$

Now, what your teacher probably wanted you to learn was the following "rule":

$$ |A \cup B| = |A| + |B| - |A \cap B|. $$

This is easy to see it is true, since to count the number of elements that are in either $A$ or $B$ (or possibly both $A$ and $B$), you count the number of elements in $A$, add to it the number of elements in $B$, and then subtract the stuff you double counted, which is precisely the elements in $A \cap B$.

Therefore, if you want to find the number of sides of a die that are (say) even or prime, you count the number of sides which are even (there are $3$ such sides - namely $2$ , $4$, $6$) add to it the number of sides which are prime (again, there are $3$ such sides - namely $2$, $3$, $5$), and then subtract the sides which we double counted (we counted the side with the number $2$ twice).

Therefore, there are $3 + 3 - 1 = 5$ sides of a die which are even or prime.

Now, you can take this strategy and count the number of cards in a deck which are either non face cards or clubs.

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