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Combinatorics/Permutations - Stuck On This Q! :-( Watch

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    8 Cards selected with replacement from a pack of 52. 12 picture cards, 20 odd cards, 20 even cards.

    a) How many different sequences of 8 cards are possible?

    I did this right: 52 choices in card 1, 52 in card 2 ... 52 in card 8
    => No. Sequences = 52^8

    b) How many of these sequences contain 3 picture cards, 3 odd cards and 2 even cards?
    I've had many goes at this, but can't get it right.

    Thanks for all your help.
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    Probability of getting such a set multiplied by the degeneracy of that set...tried that?

    ie, if I have 2 children, what are the odds that I have a son and a daughter?

    well the possible combinations are: ss, sd, ds, dd....so (1/4)*2 = 1/2
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    seem to have misread the question, but concept is still the same:

    think about the simplest case of a sequence of 8 cards:

    pic,pic,pic,odd,odd,odd,even,eve n

    how many of these are there?

    12*12*12*20*20*20*20*20.

    now how many unique permutations of p,p,p,o,o,o,e,e are there? once you figured this out it should become clear
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    (Original post by Plutoniummatt)
    Probability of getting such a set multiplied by the degeneracy of that set...tried that?
    Probability of such a set multiplied by 52^8 should give me the answer, if that's what you mean?

    (Original post by Plutoniummatt)
    ie, if I have 2 children, what are the odds that I have a son and a daughter?

    well the possible combinations are: ss, sd, ds, dd....so (1/4)*2 = 1/2
    If I have a sequence of 8, what are the odds of 3 pic, 3 odd, 2 even?

    8! permutations for 8 distinct cards
    (8!)/(3!3!2!) permutations for 3 pic, 3 odd, 2 even.
    So P = 1/(3!3!2!) = 1/72
    Then (1/72)*52^8 should give me the number of sequences overall.

    But I don't think this is right, and the book disagrees with me too.
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    the answer is 8c3*5c3*(20^3)*(20^2)*(12^3)

    Imagine there are 8 cards. 3 even, 3 odd, 2 pictures. Imagine also that all even, odd and picture cards are the same. Then the number of ways of arranging the different cards is 8C3 *5c3*2C2.


    but each picture, odd or even card is distinct. Therefore we multiply the number we just worked out by
    20^3*12^3*20^2.

    This is because 3 cards are 1 of 20 different cards, 2 cards are 1/20 different even cards and 3 cards are 1/12 different odd cards.

    I hope this is right.
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    what answer do you get? Im abit rusty on this but i seem to get:

    (8!)/(3!3!2!) multiplied by (20^5)*(12^3)

    which is about 3.09 x 10^12

    (which agrees with above poster)

    explanation:

    number of ways of arranging p,p,p,o,o,o,e,e is (8!)/(3!3!2!) I hope you get that.

    for each unique way of arranging the above combination, there are (20^5)*(12^3) ways of making them with the cards provided, multiply them together and u get the answer.
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    (Original post by Plutoniummatt)
    think about the simplest case of a sequence of 8 cards:

    pic,pic,pic,odd,odd,odd,even,eve n

    how many of these are there?

    12*12*12*20*20*20*20*20.
    Yep

    (Original post by Plutoniummatt)
    now how many unique permutations of p,p,p,o,o,o,e,e are there? once you figured this out it should become clear
    12*12*12*20*20*20*20*20 possible sequences for the case pppooee (in that order)
    8!/(3!3!2!) distinct arrangements of pppooee
    So Total Number of Sequences (3p, 3o, 2e) = (12^3.20^5).(8! / 3!3!2!)
    And the book agrees.
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    (Original post by Plutoniummatt)
    what answer do you get? Im abit rusty on this but i seem to get:

    (8!)/(3!3!2!) multiplied by (20^5)*(12^3)
    Yep! Book agrees

    (Original post by Plutoniummatt)
    explanation:

    number of ways of arranging p,p,p,o,o,o,e,e is (8!)/(3!3!2!) I hope you get that.

    for each unique way of arranging the above combination, there are (20^5)*(12^3) ways of making them with the cards provided, multiply them together and u get the answer.
    Yep. I get it. Thanks for all your help.
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    too easy next :P
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    (Original post by methusaleh)
    too easy next :P
    lol. It's easy when you can do it. Unfortunately for me, I'm a little inconsistent sometimes. I can crack hard problems and also trip up on some easy ones.

    Thanks for your help too.
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    (Original post by Physics Enemy)
    lol. It's easy when you can do it. Unfortunately for me, I'm a little inconsistent sometimes. I can crack hard problems and also trip up on some easy ones.

    Thanks for your help too.
    Thanks i just wanted some acknowledgment really lol.
 
 
 
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