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Thread: trinom

  1. #1
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    trinom

    How many terms are there in the expansion of (x+y+z)^20?
    a+b+c=20
    We have 3 toys and 20children and by using the formula:C(22,2).

    Let's suppose that we are lost in the jungle. We don't have a compass. We don't knoow any formula. How can we solve that?
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  2. #2
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    Re: trinom

    Huh?

    Edit: oh, I get what you are asking. Imagine it like this. You have three toys, so you need two separators to choose which kids will play with each toy. Line them up, and every ordering translates into a distribution of children. All children to the left of the first divider play with the first toy. All kids between the dividers play with the second toy. All to the right of the dividers play with the third toy.

    You have 20 kids plus 2 separators. So that is 20+2=22 positions. Choose two positions to be separators, the rest are children. Hence $\dbinom{22}{2}$
    Last edited by SlipEternal; Jun 2nd 2017 at 02:14 PM.
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  3. #3
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    Re: trinom

    Think that you don't know any formula and saw that question. How can we solve it just by logical thinking?
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  4. #4
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    Re: trinom

    Quote Originally Posted by kastamonu View Post
    How many terms are there in the expansion of (x+y+z)^20?
    By the binomial theorem (a+ b)^n has n terms. (x+ (y+z))^20, counting "y+ z" as a single term, has 20 terms. But then, expanding (y+ z)^n, with n going from 1 to 20, we have 1+ 2+ 3+ 4+ ...+ 20= 20(21)/2= 210 terms.

    a+b+c=20
    Yes, this is the same as asking "how many non-negative sets of (a, b, c) are there that add to 20".

    We have 3 toys and 20children and by using the formula:C(22,2).
    By using that formula what? Are you thinking of the "a", "b", "c" as the three toy and the "20th power" as the 20 children? C(22, 2)= \frac{22!}{2! 20!}= \frac{21*20}{2}= 21*10= 210 as I said before.

    Let's suppose that we are lost in the jungle. We don't have a compass. We don't know any formula. How can we solve that?
    Solve what? Getting out of the jungle? I doubt there is any "formula" for "getting out of the jungle"! I would probably scream really, really loud!
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  5. #5
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    Re: trinom

    Quote Originally Posted by kastamonu View Post
    Think that you don't know any formula and saw that question. How can we solve it just by logical thinking?
    I edited my post and explained it to you.
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  6. #6
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    Re: trinom

    Quote Originally Posted by SlipEternal View Post
    I edited my post and explained it to you.
    Many Thanks.
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  7. #7
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    Re: trinom

    Quote Originally Posted by HallsofIvy View Post
    By the binomial theorem (a+ b)^n has n terms. (x+ (y+z))^20, counting "y+ z" as a single term, has 20 terms. But then, expanding (y+ z)^n, with n going from 1 to 20, we have 1+ 2+ 3+ 4+ ...+ 20= 20(21)/2= 210 terms.


    Yes, this is the same as asking "how many non-negative sets of (a, b, c) are there that add to 20".


    By using that formula what? Are you thinking of the "a", "b", "c" as the three toy and the "20th power" as the 20 children? C(22, 2)= \frac{22!}{2! 20!}= \frac{21*20}{2}= 21*10= 210 as I said before.


    Solve what? Getting out of the jungle? I doubt there is any "formula" for "getting out of the jungle"! I would probably scream really, really loud!
    By the binomial theorem (a+b)^n has n+1 terms.
    (x+y)^2 has 3 terms.
    But 210 is true. How did you get it? How did you find 20 terms?
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  8. #8
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    Re: trinom

    Quote Originally Posted by kastamonu View Post
    By the binomial theorem (a+b)^n has n+1 terms.
    (x+y)^2 has 3 terms.
    But 210 is true. How did you get it? How did you find 20 terms?
    Lets play off of Prof. Halls post.
    Multiply out $(x+y+z)(x+y+z)(x+y+z)(x+y+z)(x+y+z)$. How is that done?
    Well select a term from each of the five factors.
    We get $Cx^ky^mz^n$ where $0\le k,n,m\le 5\;\;,k+m+n=5,$ and $C= \dfrac {5!}{(k!)(m!)(n!)}$ trinom-ex1.gif.

    Like wise $(x+y+z)^{20}$ is made up of a sum of terms: $Cx^ky^mz^n$ where $0\le k,n,m\le 20\;\;,k+m+n=20,$ and $C= \dfrac {20!}{(k!)(m!)(n!)}$
    trinom-wolframalpha6.gif.
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