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I am stuck with the question, In how many ways can you choose,
a 9, a red card with a value > 9
or a black card with a value < 6,
from a deck of cards . Now my friend solved it as follows, 4 + 10 + 8 = 22 ways.
But I think the answer should be,
52C4 + 52C10 + 52C8.
Can anyone please help me out with it?

Thanks

  • 3
    Are you drawing just a single card and wanting to get one of the three outcomes, or are you drawing a hand of some other size and hoping to get all the outcomes satisfied at once?2011-10-23
  • 1
    Aren't there only two black 6's in a deck? Where does 8 come from? Presumably you count the Aces above 9.2011-10-23
  • 0
    Yes the value of Ace is greater than all the others.2011-10-23
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    if you are drawing a single card there are 16 cases that satisfies your conditions...2011-10-23
  • 1
    But where does the 8 come from? Should that be "a black card with value **less** than 6" instead of "a black card with a value 6"?2011-10-23
  • 0
    @ArturoMagidin:Sorry, my fault...2011-10-23

1 Answers 1

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The answer $\binom{52}{4} + \binom{52}{10} + \binom{52}{8}$ is incorrect.

It seems the argument is "there are four $9$s, there are 10 cards red cards with value greater than 9, there are 8 black cards with value less than 6, so I need to choose 4 from among the 52 cards, or 10 from among the 52 cards, or 8 from among the 52 cards".

That line of thought is incorrect: $\binom{52}{4}$ tells you all the ways to pick four cards from the deck, it does not tell you how many ways you have to pick a 9. Likewise for the other summands.

Your friend is correct. Since the three categories are mutually exclusive (no card can simultaneously be two of "a nine", "a red card with value greater than 9", and "a black card with value less than 6"), all you need to do is count how many cards there are in each category and add them all up.

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    Happy 80k :-)${}$2011-10-23
  • 0
    @Asaf: Thanks! I look forward to your 16K party.2011-10-23
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    @ArturoMagidin: Are we picking 1 card at a time?2011-10-24
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    @Akito: That's what your problem seems to say.2011-10-24