Maximizing the trinomial coefficient












6












$begingroup$


Let $binom{n} {a,b,c} = frac{n!}{a!cdot b! cdot c!}$



In other words, $binom{n} {a,b,c}$ is the trinomial coefficient.



I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.



I have determined, by simply plugging in numbers, that:



When $nbmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = frac{n}{3}$



When $n bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = frac{(n - 1)}{3} space text{and} space c = frac{(n + 2)}{3} impliedby$ Note for this case, any two variables can equal $frac{(n - 1)}{3}$ while the other equals $frac{(n + 2)}{3}.$



And when $n bmod 3 = 2$, the trinomial coefficient is maximized when $a = frac{(n - 2)}{3}$ and $b = c = frac{(n + 1)}{3} impliedby$ Same caveat about the interchanging of variables as $n bmod 3 = 1$.



While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?






binomial-theorem multinomial-coefficients







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  • $begingroup$
    Do you know how to prove the analogous fact about binomial coefficients?
    $endgroup$
    – Qiaochu Yuan
    Nov 6 '14 at 0:39
















6












$begingroup$


Let $binom{n} {a,b,c} = frac{n!}{a!cdot b! cdot c!}$



In other words, $binom{n} {a,b,c}$ is the trinomial coefficient.



I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.



I have determined, by simply plugging in numbers, that:



When $nbmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = frac{n}{3}$



When $n bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = frac{(n - 1)}{3} space text{and} space c = frac{(n + 2)}{3} impliedby$ Note for this case, any two variables can equal $frac{(n - 1)}{3}$ while the other equals $frac{(n + 2)}{3}.$



And when $n bmod 3 = 2$, the trinomial coefficient is maximized when $a = frac{(n - 2)}{3}$ and $b = c = frac{(n + 1)}{3} impliedby$ Same caveat about the interchanging of variables as $n bmod 3 = 1$.



While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?






binomial-theorem multinomial-coefficients







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know how to prove the analogous fact about binomial coefficients?
    $endgroup$
    – Qiaochu Yuan
    Nov 6 '14 at 0:39














6












6








6


0



$begingroup$


Let $binom{n} {a,b,c} = frac{n!}{a!cdot b! cdot c!}$



In other words, $binom{n} {a,b,c}$ is the trinomial coefficient.



I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.



I have determined, by simply plugging in numbers, that:



When $nbmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = frac{n}{3}$



When $n bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = frac{(n - 1)}{3} space text{and} space c = frac{(n + 2)}{3} impliedby$ Note for this case, any two variables can equal $frac{(n - 1)}{3}$ while the other equals $frac{(n + 2)}{3}.$



And when $n bmod 3 = 2$, the trinomial coefficient is maximized when $a = frac{(n - 2)}{3}$ and $b = c = frac{(n + 1)}{3} impliedby$ Same caveat about the interchanging of variables as $n bmod 3 = 1$.



While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?






binomial-theorem multinomial-coefficients







share|cite|improve this question











$endgroup$




Let $binom{n} {a,b,c} = frac{n!}{a!cdot b! cdot c!}$



In other words, $binom{n} {a,b,c}$ is the trinomial coefficient.



I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.



I have determined, by simply plugging in numbers, that:



When $nbmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = frac{n}{3}$



When $n bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = frac{(n - 1)}{3} space text{and} space c = frac{(n + 2)}{3} impliedby$ Note for this case, any two variables can equal $frac{(n - 1)}{3}$ while the other equals $frac{(n + 2)}{3}.$



And when $n bmod 3 = 2$, the trinomial coefficient is maximized when $a = frac{(n - 2)}{3}$ and $b = c = frac{(n + 1)}{3} impliedby$ Same caveat about the interchanging of variables as $n bmod 3 = 1$.



While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?






binomial-theorem multinomial-coefficients




binomial-theorem multinomial-coefficients






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edited Jan 3 at 11:29









E.Nole

301114




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asked Nov 6 '14 at 0:26









user3746892user3746892

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616












  • $begingroup$
    Do you know how to prove the analogous fact about binomial coefficients?
    $endgroup$
    – Qiaochu Yuan
    Nov 6 '14 at 0:39


















  • $begingroup$
    Do you know how to prove the analogous fact about binomial coefficients?
    $endgroup$
    – Qiaochu Yuan
    Nov 6 '14 at 0:39
















$begingroup$
Do you know how to prove the analogous fact about binomial coefficients?
$endgroup$
– Qiaochu Yuan
Nov 6 '14 at 0:39




$begingroup$
Do you know how to prove the analogous fact about binomial coefficients?
$endgroup$
– Qiaochu Yuan
Nov 6 '14 at 0:39










1 Answer
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$begingroup$

Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.



Suppose that we have $xlt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
$$(x+1)!(y-1)!=frac{x+1}{y}(x!y!).$$
Thus $(x+1)!(y-1)!lt x!y!$ unless $y=x+1$, in which case we have equality.



So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.



Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.






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    $begingroup$

    Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.



    Suppose that we have $xlt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
    $$(x+1)!(y-1)!=frac{x+1}{y}(x!y!).$$
    Thus $(x+1)!(y-1)!lt x!y!$ unless $y=x+1$, in which case we have equality.



    So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.



    Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.



      Suppose that we have $xlt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
      $$(x+1)!(y-1)!=frac{x+1}{y}(x!y!).$$
      Thus $(x+1)!(y-1)!lt x!y!$ unless $y=x+1$, in which case we have equality.



      So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.



      Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.



        Suppose that we have $xlt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
        $$(x+1)!(y-1)!=frac{x+1}{y}(x!y!).$$
        Thus $(x+1)!(y-1)!lt x!y!$ unless $y=x+1$, in which case we have equality.



        So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.



        Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.






        share|cite|improve this answer











        $endgroup$



        Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.



        Suppose that we have $xlt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
        $$(x+1)!(y-1)!=frac{x+1}{y}(x!y!).$$
        Thus $(x+1)!(y-1)!lt x!y!$ unless $y=x+1$, in which case we have equality.



        So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.



        Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 6 '14 at 1:17

























        answered Nov 6 '14 at 1:04









        André NicolasAndré Nicolas

        455k36432819




        455k36432819






























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