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Math Help - Answer check data management

  1. #1
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    Answer check data management

    46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

    Answer: 5! = 5 x 4 x 3 x 2 x 1
    = 120 different words can be formed

    b) In how many of these words are M and R together?

    Answer:
    0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
    1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
    2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
    3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

    Therefore there are 48 words where M and R will be together.


    c) What is the probability of M and R not occurring together?

    Answer:
    48/120 = x/100
    = 40% probability that they will be together

    Therefore, there is a 60% probability that M and R won't be together
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  2. #2
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    Quote Originally Posted by cnmath16 View Post
    46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

    Answer: 5! = 5 x 4 x 3 x 2 x 1
    = 120 different words can be formed

    b) In how many of these words are M and R together?

    Answer:
    0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
    1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
    2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
    3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

    Therefore there are 48 words where M and R will be together.


    c) What is the probability of M and R not occurring together?

    Answer:
    48/120 = x/100
    = 40% probability that they will be together

    Therefore, there is a 60% probability that M and R won't be together
    Yes, all very good. Apart from in c), I'd just like to point out that probability of an event is usually expressed as a number between 0 and 1. Not a percentage.
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  3. #3
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    Quote Originally Posted by cnmath16 View Post
    46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

    Answer: 5! = 5 x 4 x 3 x 2 x 1
    = 120 different words can be formed

    b) In how many of these words are M and R together?

    Answer:
    0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
    1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
    2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
    3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

    Therefore there are 48 words where M and R will be together.


    c) What is the probability of M and R not occurring together?

    Answer:
    48/120 = x/100
    = 40% probability that they will be together

    Therefore, there is a 60% probability that M and R won't be together
    (b) can also be done by considering M and R as a single unit and arranging the 4 objects [MR], A, C, H to get 4! = 24.

    Then note that within the unit [MR] the M and R can arrange in 2! = 2 ways.

    So you get (24)(2) = 48.
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