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Math Help - What is the difference between Selections and Arrangements/Permutaions?

  1. #1
    Junior Member 22upon7's Avatar
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    Probability Application help

    Hi,

    I'm revising for my Maths exam on Tuesday and I really need help with this question:

    Three letters are chosen at random from the word "HEART" and arranged in a row. Find the probability that:

    a. the letter H is first
    b. the letter H is chosen
    c. both vowels are chosen


    I think this is how you work a. out:

    HEART has 5 letters, so 1/5 = 0.2

    I'm not sure how to work b. and c. out though, I think it needs permutation and/or combinations but I can't work out the correct answer.

    Any help would be greatly appreciated, and I'm sure I'll have more questions to come.

    Thanks,

    Dru
    Last edited by 22upon7; November 5th 2009 at 10:37 PM.
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  2. #2
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    Three letters are chosen at random from the word "HEART" and arranged in a row. Find the probability that:

    a. the letter H is first

    So, three letters are chosen randomly and arranged in a row. There are 5P3=60 ways they can be arranged... Since you want the first letter to be a H, there is only one way that can happen, and there is 4P2=12 ways that can happen. So Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}

    b. the letter H is chosen
    Letter H can now be in 1 of 3 places, so
    P(\mbox{Letter H is chosen})=\frac{3\cdot 12}{60}=\frac{3}{5}

    c. both vowels are chosen
    So there is 2P2=2 ways to arrange the 2 vowels, and the remaining 3 characters can be arranged in any of the 3 positions, so there are 9 ways for the other letter to be positioned.
    So P(\mbox{both vowels are chosen})=\frac{2\cdot 9}{60}=\frac{3}{10}
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  3. #3
    Junior Member 22upon7's Avatar
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    Quote Originally Posted by Robb View Post
    So, three letters are chosen randomly and arranged in a row. There are 5P3=60 ways they can be arranged... Since you want the first letter to be a H, there is only one way that can happen, and there is 4P2=12 ways that can happen. So Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}
    Thanks Robb, I didn't understand this bit though, could you help me out with it?:

    Quote Originally Posted by Robb View Post
    Since you want the first letter to be a H, there is only one way that can happen, and there is 4P2=12 ways that can happen.
    Thanks again,

    Dru
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  4. #4
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    Just the permuation formula,
    \frac{4!}{(4-2)!}=\frac{24}{2}=12
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  5. #5
    Junior Member 22upon7's Avatar
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    yes, but how did you get those values? The 4 and the 2?

    Thanks again,

    Dru
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  6. #6
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    Oh, sorry. Orignally you had 5 letters to choose from, and you were selecting 3. But because the first one is a H, you are now selecting 2 letters from the remaining 4.
    So the probability is just the ways you can arrange H with 2 other letters, divided by the total number of ways you could arrange 3 letters from 5
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  7. #7
    Junior Member 22upon7's Avatar
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    No worries, Thanks again mate!
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