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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?


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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? – Austin Mohr Oct 23 '11 at 17:06
Aren't there only two black 6's in a deck? Where does 8 come from? Presumably you count the Aces above 9. – Ross Millikan Oct 23 '11 at 17:11
Yes the value of Ace is greater than all the others. – Fahad Uddin Oct 23 '11 at 17:19
if you are drawing a single card there are 16 cases that satisfies your conditions... – pedja Oct 23 '11 at 17:33
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"? – Arturo Magidin Oct 23 '11 at 19:33
up vote 4 down vote accepted

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 :-)${}$ – Asaf Karagila Oct 23 '11 at 20:07
@Asaf: Thanks! I look forward to your 16K party. – Arturo Magidin Oct 23 '11 at 20:09
@ArturoMagidin: Are we picking 1 card at a time? – Fahad Uddin Oct 24 '11 at 5:36
@Akito: That's what your problem seems to say. – Arturo Magidin Oct 24 '11 at 13:01

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