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Math Help - permutation help

  1. #1
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    permutation help

    hi, im currently student in vietnam

    i have 2 permutation questions

    1. how many words consisitng of 3 vowel and 2 consonants can be formed from the letters of the word "columbians"

    2. a presidnet secretry and treasurer are to be named from 5 men and 3 women
    how many permutations are there if one man and one women must be picked

    heres how much i have done


    1) you have 4 vowels and 6 consonants and 10 letters total

    you need a 5 letter permutation

    so you have 4 x 3 x 2 x 6 x 5

    but if you do this, will you always have the order in 3 vowels and 2 consonants? what do i multiply this by to mix it up?

    keep in mind my total permutations without limits is 10!/5!


    2) so i do the same thing as above, i know 5 men, 3 women and a total of 8 ppl

    5 x 3 x 6

    but wont the man always be president? again, how do i mix this up??



    sorry if my english not very good
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  2. #2
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    Quote Originally Posted by gogeta View Post
    1. how many words consisitng of 3 vowel and 2 consonants can be formed from the letters of the word "columbians"
    Since you did not say, we assume the letters cannot be repeated.
    \dbinom{4}{3}\dbinom{6}{2}(5!)
    Choose the vowels, choose the two consonants, then arrange the five.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Since you did not say, we assume the letters cannot be repeated.
    \dbinom{4}{3}\dbinom{6}{2}(5!)
    Choose the vowels, choose the two consonants, then arrange the five.
    yeah but your answer is 86400, which is greater than

    P(10,5) which is if I picked 5 random letters from the 10 in COLUMBIANS
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  4. #4
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    You need to learn to do calculations:
    \dbinom{4}{3}\dbinom{6}{2}(5!)=7200
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  5. #5
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    Isn't this a permutation question because the order that you place the letters in matters...?
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  6. #6
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    im unclears as to why you use combinations instead of permutation
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  7. #7
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    Quote Originally Posted by gogeta View Post
    im unclears as to why you use combinations instead of permutation
    You must first choose three vowels \dbinom{4}{3}=4
    Then choose the two consonants \dbinom{6}{2}=15
    Now you have five to arrange.
    This is a mixed problem.
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  8. #8
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    Quote Originally Posted by Plato View Post
    You must first choose three vowels \dbinom{4}{3}=4
    Then choose the two consonants \dbinom{6}{2}=15
    Now you have five to arrange.
    This is a mixed problem.
    thank you i see now that it doesn't matter which vowels you pick, b/c you mix them up by doing 5!
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  9. #9
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    how would i be able to do the 2nd question, do i multiply by 3!

    but then that would be more than P(18,3)
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  10. #10
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    Quote Originally Posted by gogeta View Post
    how would i be able to do the 2nd question
    I will give the answer and expect you to explain it.
    ^N\mathcal{P}_k is the permutation of N taken k at a time.
    ANSWER : ^8\mathcal{P}_3~-^5\mathcal{P}_3-~^3\mathcal{P}_3 .
    WHY?
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  11. #11
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    Quote Originally Posted by Plato View Post
    I will give the answer and expect you to explain it.
    ^N\mathcal{P}_k is the permutation of N taken k at a time.
    ANSWER : ^8\mathcal{P}_3~-^5\mathcal{P}_3-~^3\mathcal{P}_3 .
    WHY?
    8 permutate 3 is all possible

    5P3 is all men
    3P3 is all women

    left with all other combos since we have subtracted all the ones that are not 2men1female or 2female1man
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  12. #12
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    Exactly.
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  13. #13
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    Quote Originally Posted by Plato View Post
    Exactly.
    just out of curiousity is it possible to do this question the way I have originally tried to do it in my opening post?
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