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Math Help - Arrangements and combinations

  1. #1
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    Arrangements and combinations

    1.In a true false test there are 12 questions. If a student decides to check six of each at random, in how many ways can she do it.

    (I was thinking the answer would be 12 choose 6 or 12!/6!) Can anybody explain there reasoning in arriving to their answer.

    2.in how many ways can a committee of 3 be chosen from 20 people? In how many ways can a president, a secretary and a treasurer be chosen?

    (For the second part of this question i was thinking it was 20!, 19!, 18!)

    Thanks again for any replies.
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  2. #2
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    Quote Originally Posted by multivariablecalc View Post
    1.In a true false test there are 12 questions. If a student decides to check six of each at random, in how many ways can she do it.
    I do not understand what is meant by “a student decides to check six of each at random”
    Does it mean “how many ways can she chose 6 questions to answer?”
    Or does it mean “how many ways can she answer 6 questions?”
    Or does it mean “how many ways can she chose and then answer 6 questions?”


    Quote Originally Posted by multivariablecalc View Post
    2.in how many ways can a committee of 3 be chosen from 20 people? In how many ways can a president, a secretary and a treasurer be chosen?
    How many ways can a committee of 3 be chosen from 20 people: \binom{20}{3}=\frac{20!}{3!\cdot 17!}.

    In how many ways can a president, a secretary and a treasurer be chosen? \frac{20!}{17!}
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  3. #3
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    Hi Plato thanks for the reply. For the first problem i think the wording is really what has confused me for this question. I typed it verbatim out of the book. The Answer is (12 6) or a set of 12 choose 6.

    For the second part of the second question why would be 17!? wouldn;t it be 20!19!18! becuase you have one person left to choose from? Thanks again for your response.
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  4. #4
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    Quote Originally Posted by multivariablecalc View Post
    For the second part of the second question why would be 17!? wouldn;t it be 20!19!18! becuase you have one person left to choose from? Thanks again for your response.
    20!=(20)(19)(18)(17!)
    So the answer is just 20\cdot 19\cdot 18
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