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  1. #1
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    New Question

    Hey again !
    1) We have for example, 100 Questions to be answered for a test (Yes/No).
    In order to pass the test you must have at least 95 right answers. So, how many ways there exist in order someone to pass this exam ?

    I believe that firstly we must find out, how many different combinations of 5digits (made by numbers from 1 to 100) we have, and then we just add to this, +5 cases ( 96(+1), 97(+2), 98(+3), 99(+4), or 100(+5) right answers).

    But actually my problem is how to present that factorial, given that my rumination is right.


    2) For Example, we have this word --> SUCCESS
    How many different words (not needed to be in the lexicon) can be made if we change the order of the letters?

    My thought is: 7! / 3!2! = 420 new words right?

    Thanks in advance for another time! :=)
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  2. #2
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    Your answer is right. The letters can be arranged in 7! distinct ways. And "c" occurs twice and "s" 3 times, so you have to divide out the letters that occur more than once \frac{7!}{2! 3!}.
    Last edited by Roam; January 15th 2010 at 12:39 PM.
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  3. #3
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    Your 2) is quite right!

    7 letters gives 7! arrangements,
    but unless we colour the C's, we cannot distinguish between them
    and the same for the S's.
    Hence there would be 2! as many distinct arrangements if the C's were different, along with 3! as many arrangements if the S's were different.
    Hence if all the letters were different, there would be 2!3! times as many arrangements.

    For 1) there is only \binom{100}{100}=1 way to get 100 correct.

    There are \binom{100}{99} ways to get 99 correct,
    or \binom{100}{1}=100 ways to get 1 wrong.

    There are \binom{100}{98} ways to get 98 right,
    or \binom{100}{2} ways to get 2 wrong.

    Of course \binom{100}{98}=\binom{100}{2}

    Continue on like this to find out how many you have for each case.
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  4. #4
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    Archie Meade, Thanks again for your response!

    So here is what i 've done.

    IF we have 100 Questions to be answered for a test (Yes/No),
    and we must give at least 95 right answers to pass the exam, we have the following ways to pass it :

    <br /> <br />
\binom{100}{100}+\binom{100}{99}+\binom{100}{98}+\  binom{100}{97}+\binom{100}{96}+\binom{100}{95}

    = 1+ 100+ 4950+ 161700+ 3921225+ 75287519 =79375495


    Is that right ?

    Thanks again =)
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  5. #5
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    It may appear to be a large number,
    but yes, that's it, though can you double-check, as i get 75287520
    for the number of ways to get 95 right answers,
    or 5 wrong answers.

    Notice that you get the exact same answer
    by counting the corresponding "wrong answers".

    \binom{100}{0}+\binom{100}{1}+\binom{100}{2}+\bino  m{100}{3}+\binom{100}{4}+\binom{100}{5}
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