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Oct 17, 2019
 #1
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Oct 17, 2019
 #15
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n chairs around a circular table, m people, at least k chairs between them (n>=m(k+1))

 

If chairs are numbered, people are distinguishable:

 

\(n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}=a(n,m,k)\)

 

If chairs are numbered, people are indistinguishable:

 

\(\frac{n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}}{m!}=\frac{n*{n-m*(k+1)+(m-1) \choose m-1}}{m}=b(n,m,k)\)

 

If chairs are not numbered (indistinguishable), people are distinguishable:

 

\(\frac{n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}}{n}=(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}=c(n,m,k)\)

 

However if the chairs are indistinguishable AND the people are indistinguishable, we can't always divide the third formula by m factorial (or alternatively, the second one by n the number of chairs) to get the number of arrangements where both chairs and people are indistinguishable.

 

the formula is (using burnside's lemma):

\(\frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} b(gcd(i,n), \frac{m}{\frac{n}{gcd(i,n)}}, k) = \frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{gcd(i,n)*{gcd(i,n)- \frac{m}{\frac{n}{gcd(i,n)}}(k+1)+(\frac{m}{\frac{n}{gcd(i,n)}}-1)\choose (\frac{m}{\frac{n}{gcd(i,n)}}-1)}}{\frac{m}{\frac{n}{gcd(i,n)}}} \\=\frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{n*{gcd(i,n)- \frac{gcd(i,n)*m}{n}(k+1)+ \frac{gcd(i,n)*m}{n}-1)\choose ( \frac{gcd(i,n)*m}{n}-1)}}{m}\)

 

 

so we can divide the second formula to get the last one if and only if gcd(n,m)=1

Oct 17, 2019
 #4
avatar+26367 
+3

Let x,y, and z be positive real numbers such that \(\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}\) = 1.
Find the minimum value of \(\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}\).

 

1.  \(x^4 y^4 + x^4 z^4 + y^4 z^4 = \ ?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ 1 &=& \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^4 y^4 z^4} \\\\ \mathbf{x^4 y^4 + x^4 z^4 + y^4 z^4} &=& \mathbf{ x^4 y^4 z^4 } \\ \hline \end{array}\)

 

2. rearrange \(\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}\)

\(\begin{array}{|rcll|} \hline \dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} &=& \dfrac{x^4 y^4 z^4 } {x^3 y^2 z^3} \\\\ \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3} } &=& \mathbf{x y^2 z} \\ \hline \end{array}\)

 

Lagrange Multipliers

\(\begin{array}{|rcll|} \hline g(x,y,z) &=& \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 = 0 \\\\ f(x,y,z) &=& xy^2z \rightarrow \text{MIN.} \\\\ L(x,y,z,\lambda) &=& \underbrace{xy^2z}_{f(x,y,z)} + \lambda \left( \underbrace{\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4} - 1 }_{g(x,y,z)}\right) \\\\ \hline \frac{\partial L(x,y,z,\lambda)}{\partial x} &=& y^2z-\dfrac{4}{x^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial y} &=& 2xyz-\dfrac{4}{y^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial z} &=& xy^2-\dfrac{4}{z^5} \lambda \\\\ \frac{\partial L(x,y,z,\lambda)}{\partial \lambda} &=& 0+ \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline & y^2z-\dfrac{4}{x^5} \lambda &=& 0 \quad | \quad \cdot x^5 \\\\ (1) & y^2x^5z-4\lambda &=& 0 \\ \hline & 2xyz-\dfrac{4}{y^5} \lambda &=& 0 \quad | \quad \cdot y^5 \\\\ (2) & 2xy^6z-4\lambda &=& 0 \\ \hline & xy^2-\dfrac{4}{z^5} \lambda &=& 0 \quad | \quad \cdot z^5 \\\\ (3) & xy^2z^5-4\lambda &=& 0 \\ \hline (4) & \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline & y^2x^5z-4\lambda &=& 0 \qquad (1) \\ (5) & \mathbf{z} &=& \mathbf{ \dfrac{4\lambda}{y^2x^5} } \\ \hline \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \\ (7) & \mathbf{x} &=& \mathbf{ \dfrac{4\lambda}{2y^6z} } \\ \hline & 2xy^6z-4\lambda &=& 0 \qquad (2) \quad | \quad \mathbf{z=\dfrac{4\lambda}{y^2x^5} } \\ & 2xy^6\dfrac{4\lambda}{y^2x^5}-4\lambda &=& 0 \\ & 2y^4\dfrac{4\lambda}{x^4} &=& 4\lambda \\ & 2y^4 &=& x^4 \\ (8)& \mathbf{y^4} &=& \mathbf{\dfrac{x^4}{2}} \\ \hline & xy^2z^5-4\lambda &=& 0 \qquad (3) \quad | \quad \mathbf{x=\dfrac{4\lambda}{2y^6z} } \\ & \dfrac{4\lambda}{2y^6z}y^2z^5-4\lambda &=& 0 \\ & \dfrac{4\lambda}{2y^4}z^4 &=& 4\lambda \\ & z^4 &=& 2y^4 \\ (9) & \mathbf{y^4} &=& \mathbf{ \dfrac{z^4}{2} } \\ \hline (8)=(9): & y^4 = \dfrac{x^4}{2} &=& \dfrac{z^4}{2} \\\\ (10) & \mathbf{x^4} &=& \mathbf{z^4} \\ \hline \end{array} \)

 

\(\mathbf{z=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \qquad (4) \quad | \quad \mathbf{x^4=z^4} \\\\ \dfrac{1}{z^4} + \dfrac{1}{y^4} + \dfrac{1}{z^4}-1 &=& 0 \quad | \quad \mathbf{y^4=\dfrac{z^4}{2}} \\\\ \dfrac{1}{z^4} + \dfrac{1}{\dfrac{z^4}{2}} + \dfrac{1}{z^4}-1 &=& 0 \\\\ \dfrac{1}{z^4} + \dfrac{2}{z^4} + \dfrac{1}{z^4} &=& 1 \\\\ \dfrac{4}{z^4} &=& 1 \\\\ z^4 &=& 4 \\ z^2 &=& 2 \\ \mathbf{z} &=& \mathbf{\sqrt{2}} \\ \hline \end{array} \)

 

\(\mathbf{x=\ ?} \)

\(\begin{array}{|rcll|} \hline \mathbf{x^4} &=& \mathbf{z^4} \qquad (10) \quad | \quad \mathbf{z^4 = 4} \\\\ x^4 &=& 4 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

 

\(\mathbf{y=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{y^4} &=& \mathbf{\dfrac{z^4}{2} } \qquad (9) \quad | \quad \mathbf{z^4 = 4} \\ y^4 &=& \dfrac{4}{2} \\ y^4 &=& 2 \\ y &=& \sqrt[4]{2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x^4 y^4 + x^4 z^4 + y^4 z^4}{x^3 y^2 z^3}} &=& xy^2z \\\\ &=& \sqrt{2} \left( \sqrt[4]{2} \right)^2 \sqrt{2} \\ &=& 2 \cdot \left( \sqrt[4]{2} \right)^2 \\ &=& 2\cdot 2^{\frac{2}{4}} \\ &=& 2\cdot 2^{\frac{1}{2}} \\ &=& \mathbf{2 \sqrt{2}} \\ \hline \end{array}\)

 

 

laugh

Oct 17, 2019
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Oct 17, 2019
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Oct 17, 2019

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