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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

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    @ArturoMagidin:Sorry, my fault...2011-10-23

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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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    @Akito: That's what your problem seems to say.2011-10-24