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Difficult Math Question #20 - Combinations

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #20 - Combinations

Post Fri Sep 22, 2006 10:04 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    OA coming after some people answer..

    In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
    a. (13^4) x 48 x 47
    b. (13^4) x 27 x 47
    c. 48C6
    d. 13^4
    e. (13^4) x 48C6

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    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Oct 04, 2006 4:58 am
    i think its about time to see the answer Smile

    i think it should be < 13^4 but i didnt find an easy way to solve it

    can you tell us how to solve this?

    Thanks

    abby_g Junior | Next Rank: 30 Posts Default Avatar
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    Post Wed Oct 04, 2006 8:01 am
    THe answer is A.
    A spade is selected in 13C1 ways. such other 3 so total of (13C1)^4.

    now we want 2 cards. one card i can get in 48C1 and the other in 47C1.

    13^4 * 48 * 47.

    the 2 cards can be picked using 48P2, since we want ways. ANYWAYS the answer is still the same.

    i hope its right. Curiousity kills the cat Answer please.

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Oct 04, 2006 8:24 pm
    OA:

    Sol: 52 cards in a deck -13 cards per suit
    First card - let us say from suit hearts = 13C1 =13
    Second card - let us say from suit diamonds = 13C1 =13
    Third card - let us say from suit spade = 13C1 =13
    Fourth card - let us say from suit clubs = 13C1 =13
    Remaining cards in the deck= 52 -4 = 48
    Fifth card - any card in the deck = 48C1
    Sixth card - any card in the deck = 47C1

    Total number of ways = 13 * 13 * 13 * 13 * 48 * 47 = 13^4 *48*47 ---> choice A

    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Post Thu Oct 05, 2006 12:13 am
    but arent you repeating on options like this?

    this is what's confusing, i dont think you can just multiply (13^4) by 48*47

    lets take 1,2,3 of type hart and 4,5,6 of 3 different types, so:

    {1,4,5,6} {2,3}
    {2,4,5,6} {1,3}
    {3,4,5,6} {2,1}

    are the same choises, you have here a very large number of repetitions

    please elaborate on this if you understand what i mean

    Thanks

    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Post Thu Oct 05, 2006 12:16 am
    i think i got it ...

    the solution applies only for cases where the order of picking matters, so in this case:

    {1,2,3,4} {5,6}

    is different than

    {2,2,3,4} {3,1}

    This should be clarified in the question!, otherwise the solution should be different ... (i think Smile)

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