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    thanks for clicking depite the hellish title of the thread. ok, here goes...
    q1) a coin is tossed 10 times, how may differant sequences are possible?

    (i thought this would simply be 10P2 - but no, its 2^10 - why???????)

    q2) (same theme) 8 cards are dealt from a pack of 52 cards, with replacement, how may differant sequances are possible?

    (again its not 52P8 - its 52^8 - *cries*)

    my question basically is what makes these situations unsuitable to use permutations???
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    the first might be true if you were using a 10 sided coin

    however a coin is only two sided, so in each case you get one of 2 possiblities: 1 of 2 permutations.

    so if you do that 10 times, you have 2^10 permutations (think of a tree diagram...how many different branches would it have!?)

    For question 2 once again imagine drawing a tree diagram. After one card draw, there are 52 different branches, after two draws there are 52*52 different branches (cos each branch branches 52 times). So the total number of brances after 8 draws if 52^8

    Remember that permutations are used for selection without replacement. In the coin toss, there IS replacement, because you can toss a head again after you've already tossed a head.
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    ah, actually that makes sense! thanks alot!
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    yeh does indeed willa!

    exam will most certainly be something like ABCDEU vowels next to each other etc

    or STATISTICS number of possiblwe arrangemwentsdtf sd jsdfn argghe hands sore
 
 
 
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