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Math Help - Combinations (Anagrams)

  1. #1
    Sox
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    Combinations (Anagrams)

    How many anagrams can be made from the phrase "Apple pizza" if the "L" has to be followed by an a, e, or i?
    I think there there are P(10: 3,2,2,1,1,1) = 10! / (3! * 2! * 2!) anagrams without the L restriction. Any help with how to factor in the L restriction would be appreciated.
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    Quote Originally Posted by Sox View Post
    How many anagrams can be made from the phrase "Apple pizza" if the "L" has to be followed by an a, e, or i?
    I think there there are P(10: 3,2,2,1,1,1) = 10! / (3! * 2! * 2!) anagrams without the L restriction. Any help with how to factor in the L restriction would be appreciated.
    You are correct in the total number.
    There are two case with restrictions: la, li(e).
    That is l followed by ‘a’ or l followed by one of ‘e’ or ‘i’.

    Case I, \frac{9!}{(2!)(3!)}.
    Case II, \frac{2(9!)}{(3!)(2!)^2}.

    Can you explain both of those?
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  3. #3
    Sox
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    Re: Plato

    Thanks!
    Makes sense. We act as if there are 9 letters, one of which is the letter 'la', 'li', or 'le'
    CaseI is where the 9th letter is 'la', CaseII is where it is 'le' or 'li'. Case 1 does not have a second 2! because 'la' and 'a' are different letters. So solutions is: CaseI + 2*CaseII
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    Hello, Sox!

    I have a slightly different answer . . .


    How many anagrams can be made from the phrase "APPLE PIZZA"
    if the L has to be followed by an A, E,\text{ or }I ?

    We have 10 letters with some repetition: . PPP,AA,ZZ,E,I,L

    We have three cases to consider . . .


    LE: ,Duct-tape LE together.
    . . . .Then we have 9 "letters" to arrange: . \boxed{LE}\,, P,P,P,A,A,Z,Z,I
    . . . .There are: . \frac{9!}{3!2!2!}\,=\,15,120 possible anagrams.

    LI: .Duct-tape LI together.
    . . . .Then we have 9 "letters" to arrange: . \boxed{LI}\,, P,P,P,A,A,Z,Z,E
    . . . .There are: . \frac{9!}{3!2!2!}\,=\,15,120 possible anagrams.

    LA: .Duct-tape LA together.
    . . . .Then we have 9 "letters" to arrange: . \boxed{LA}\,, P,P,P,Z,Z,A,E,I
    . . . .There are: . \frac{9!}{3!2!} possible anagrams.

    . . . .But in each of these anagrams, the two A's can be switched.
    . . . . . without creating a new anagram.
    . . . .Our count is too large by a factor of 2.

    . . . .There are: . \frac{9!}{3!2!2} \,=\,15,120 anagrams for this case.


    Got it?

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  5. #5
    Sox
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    Re: Soroban

    Thanks for the response.

    "
    But in each of these anagrams, the two 's can be switched without creating a new anagram. Our count is too large by a factor of 2."

    I don't think this is correct. I think we need to act like there is only one A, the other is an "LA". These are two entirely different letters. You can't just switch A with A, you would have to switch A with "LA". So the third case would be as 9! / (3! * 2!)
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  6. #6
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    Quote Originally Posted by Sox View Post
    Thanks for the response.

    "
    But in each of these anagrams, the two 's can be switched without creating a new anagram. Our count is too large by a factor of 2."

    I don't think this is correct. I think we need to act like there is only one A, the other is an "LA". These are two entirely different letters. You can't just switch A with A, you would have to switch A with "LA". So the third case would be as 9! / (3! * 2!)
    You are correct about about that.
    That is the number of ways of arranging X,P,P,P,Z,Z,A,E,I where X=\boxed{LA}
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