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Thread: Permutation question: Francisco Scaramanga...

  1. #1
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    Permutation question: Francisco Scaramanga...

    The name is: Francisco Scaramanga.

    a) Determine the number of permutations of the letters in his name.
    This I understand. Its just 19!/2!5!2!2!3! which is 2.11 x 10^13.

    b) Determine the number of permutations of the letter in his name if:
    i) you must begin with a consonant
    ii) you must begin with a vowel.

    So just some basic information to make the questions easier:
    There are 12 consonants, 7 vowels.
    r repeats 2 times, a repeats 5 times, n repeats 2 times, s repeats 2 times and c repeats 3 times.

    For b)i) I get that their are 7 possibilities for the first letter and for b)ii) that there are 3 possibilities for the first letter.

    I just don't get how you would figure the number of permutations for the remaining letters for each question in b.
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    Re: Permutation question: Francisco Scaramanga...

    Hey Latinized.

    Try thinking of how many possibilities are reduced when you start with a vowel or consonant.

    Have you ever used a tree diagram before?
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    Re: Permutation question: Francisco Scaramanga...

    Quote Originally Posted by Latinized View Post
    The name is: Francisco Scaramanga.
    a) Determine the number of permutations of the letters in his name.
    This I understand. Its just 19!/2!5!2!2!3! which is 2.11 x 10^13.
    b) Determine the number of permutations of the letter in his name if:
    i) you must begin with a consonant
    ii) you must begin with a vowel.
    So just some basic information to make the questions easier:
    There are 12 consonants, 7 vowels.
    r repeats 2 times, a repeats 5 times, n repeats 2 times, s repeats 2 times and c repeats 3 times.
    For b)i) I get that their are 7 possibilities for the first letter and for b)ii) that there are 3 possibilities for the first letter.
    I just don't get how you would figure the number of permutations for the remaining letters for each question in b.
    b) i)
    We need to add-up three numbers: the rearrangements can begin three ways.
    1) with a $c$ : $\dfrac{18!}{(2!)^4(5!)}$
    2) with a $r,~s,\text{ or }n:~ 3\dfrac{18!}{(2!)^2(3!)(5!)}$
    3) with a $f,~g,\text{ or }m:~ 3\dfrac{18!}{(2!)^3(3!)(5!)}$.

    Please show your work on the next part.
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    Re: Permutation question: Francisco Scaramanga...

    Quote Originally Posted by Plato View Post
    b) i)
    We need to add-up three numbers: the rearrangements can begin three ways.
    1) with a $c$ : $\dfrac{18!}{(2!)^4(5!)}$
    2) with a $r,~s,\text{ or }n:~ 3\dfrac{18!}{(2!)^2(3!)(5!)}$
    3) with a $f,~g,\text{ or }m:~ 3\dfrac{18!}{(2!)^3(3!)(5!)}$.

    Please show your work on the next part.
    I'm confused on how all that added up gets you the possibilities for the first letter? Shouldn't those be the possibilities for the remaining letters since you have 18 left for the remaining letters. Could you also tell me the final solution with all the letters.
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    Re: Permutation question: Francisco Scaramanga...

    Quote Originally Posted by Latinized View Post
    I'm confused on how all that added up gets you the possibilities for the first letter? Shouldn't those be the possibilities for the remaining letters since you have 18 left for the remaining letters. Could you also tell me the final solution with all the letters.
    "Could you also tell me the final solution with all the letters." The answer is absolutely no. There are homework services that you can pay and get it done. If you only want an answer the go elsewhere.

    Part b, i has three parts it asks 'how many ways can a rearrangement begin a consonant?'
    If we start with a c it occurs three times.
    If we start with a r, s, or n they occur two times each.
    If we start with a f, g, or m they occur only one times each.

    If you want to learn how to do this kind of question, you will try to follow my answers.
    Adding the three gives us the total number that begin with a consonant.

    If you do not want to do that then bye-bye.
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    Re: Permutation question: Francisco Scaramanga...

    Quote Originally Posted by Plato View Post
    "Could you also tell me the final solution with all the letters." The answer is absolutely no. There are homework services that you can pay and get it done. If you only want an answer the go elsewhere.

    Part b, i has three parts it asks 'how many ways can a rearrangement begin a consonant?'
    If we start with a c it occurs three times.
    If we start with a r, s, or n they occur two times each.
    If we start with a f, g, or m they occur only one times each.

    If you want to learn how to do this kind of question, you will try to follow my answers.
    Adding the three gives us the total number that begin with a consonant.

    If you do not want to do that then bye-bye.
    I was only asking for a final solution to check if my answer is right b/c I didn't do it your way. I did (12x18!)/(2!5!3!2!2!), there are 12 constants possible, then 18 left. This got me 1.3x10^13. I was just seeing if your method may be different, but it got the same answer.
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    Re: Permutation question: Francisco Scaramanga...

    Quote Originally Posted by Latinized View Post
    I was only asking for a final solution to check if my answer is right b/c I didn't do it your way. I did (12x18!)/(2!5!3!2!2!), there are 12 constants possible, then 18 left. This got me 1.3x10^13. I was just seeing if your method may be different, but it got the same answer.
    HERE is the calculation,
    Thanks from Latinized
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