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Math Help - Counting Methods

  1. #1
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    Counting Methods

    Arggh! This counting methods thing is really getting to my head!!! PLEASE PLEASE PLEASE HELP ME!


    1.) In how many ways can a hand of six cards be dealt from a pack of 52 cards?

    For this question i drew up a diagram of 52 cars and 6 spots for the 6 cards. I reasoned that it probably is 6! but i was found to be incorrect, please correct me.

    I don't quite understand the distinction between Permutations and Combinations, take this for example:

    2.) Four runners are competing in a race, they are given respective tags of A,B,C and D. Only two prizes for the first and second racer will be distributed:

    a.) Find the different combinations of winners. I found that it could be AB, AC, AD, BC,BD and CD But i don't quite understand how i got it, could someone please explain it to me.

    b.) Find the combinations of the 1st and 2nd winners. Totally clueless for this one
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  2. #2
    Junior Member Lonehwolf's Avatar
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    Quote Originally Posted by andrew2322 View Post
    2.) Four runners are competing in a race, they are given respective tags of A,B,C and D. Only two prizes for the first and second racer will be distributed:

    a.) Find the different combinations of winners. I found that it could be AB, AC, AD, BC,BD and CD But i don't quite understand how i got it, could someone please explain it to me.

    Basically there are 4 runners, with tags from A-D. Only two of those can win, so you have to find out any and all possible combinations of the 2 winners ( in this case it's irrelevant whether they get 1st or 2nd. So just point out which two get in)

    A
    B
    C
    D

    Put all possibilities as I did above, and then start combining the rest with those above them.

    A B C D
    B C D
    C D
    D (As you can see D appears in all the above so no need to recombine it)

    Combine all bold letters with the normal ones to build up:

    AB, AC, AD, BC, BD, CD.

    Why didn't we write it like this:

    A B C D
    B A C D
    C A B D
    D A B C

    Because as you can see, BA is AB in the first row, so it's covered. Do note however that this method is definitely necessary for the second question. Since you have to point out who gets first and who gets second.



    b.) Find the combinations of the 1st and 2nd winners. Totally clueless for this one

    So, in this case, we write what I wrote just above once again:

    A
    B C D
    B A C D
    C A B D
    D A B C

    Now, the bold letters are those that came in FIRST. We have to determine who came in second, so you write them like this:

    AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC.

    As you may have noticed, all of those could come in first, followed by the second letter indicating 2nd place.

    If things are still unclear reply.
    .
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  3. #3
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    [quote=andrew2322;248291]Arggh! This counting methods thing is really getting to my head!!! PLEASE PLEASE PLEASE HELP ME!


    1.) In how many ways can a hand of six cards be dealt from a pack of 52 cards?

    For this question i drew up a diagram of 52 cars and 6 spots for the 6 cards. I reasoned that it probably is 6! but i was found to be incorrect, please correct me.

    I don't quite understand the distinction between Permutations and Combinations, take this for example:


    i think order and sequence is important in permutation, there must be no repetition.
    the question is combination.. i think the answer is 32468436..
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