# Counting Color Combinations in 12 Triangles

• MHB
• maxkor
In summary, the number of possible color combinations in 12 triangles is 2^12 = 4096. This is calculated by using the formula 2^n, where n is the number of triangles. The process of counting color combinations involves multiplying the number of choices for each triangle. There are no specific limitations or constraints on the color combinations, but additional requirements may reduce the number of possible combinations. Other mathematical formulas, such as nCr, can also be used to calculate the number of color combinations. The number of triangles directly affects the number of color combinations, as the total number of combinations can be found using the formula 2^n. As the number of triangles increases, the number of color combinations also increases exponentially.
maxkor
There are 12 triangles (picture). We color each side of the triangle in red, green or blue. Among the $3^{24}$ possible colorings, how many have the property that every triangle has one edge of each color?

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• ry.png
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[TIKZ]\coordinate (A) at (30:2) ;
\coordinate (B) at (90:2) ;
\coordinate (C) at (150:2) ;
\coordinate (D) at (210:2) ;
\coordinate (E) at (270:2) ;
\coordinate (F) at (330:2) ;
\coordinate (U) at (0:5) ;
\coordinate (V) at (60:5) ;
\coordinate (W) at (120:5) ;
\coordinate (X) at (180:5) ;
\coordinate (Y) at (240:5) ;
\coordinate (Z) at (300:5) ;
\foreach \point in {A,B,C,D,E,F,U,V,W,X,Y,Z} \fill [black] (\point) circle (3pt) ;
\draw (A) -- (B) -- (C) -- (D) -- (E) -- (F) -- (A) -- (U) -- (V) -- (W) -- (X) -- (Y) -- (Z) -- (U) --(A) -- (V) -- (B) -- (W) -- (C) -- (X) -- (D) -- (Y) -- (E) -- node[ left ]{$x$}(Z) -- node[ right ]{$z$}(F) -- node[ below ]{$y$}(U) ;
\draw (-90:3.5) node{$6$} ;
\foreach \x in {0,30,...,240} \draw (\x:3.5) node{$2$} ;
\draw (270:2.75) node{$A$} ;
\draw (240:2.75) node{$B$} ;
\draw (0:2.75) node{$C$} ;
\draw (330:2.75) node{$D$} ;
\draw (300:2.75) node{$E$} ;[/TIKZ]
Start with the triangle labelled $A$ at the bottom of the diagram. There are 6 ways to colour its three sides in different colours. Next, look at triangle $B$. One of its sides has already been coloured, and there are 2 ways to colour the remaining sides. Continuing in this way clockwise round the diagram, there are two ways to colour each of the triangles up to and including the one labelled $C$. There are now two cases to consider for the remaining triangles $D$ and $E$. If side $x$ in triangle $A$ and side $y$ in triangle $C$ have different colours then there is only one choice for the colour of side $z$ and therefore only 1 way to colour triangles $D$ and $E$. But if $x$ and $y$ have the same colour then there are two choices for the colour of side $z$, and therefore 2 ways to colour triangles $D$ and $E$.

Here's where the argument becomes probabilistic and unreliable. If the colours of $x$ and $y$ had been chosen independently at random, then the probability of them being the same would be $\frac13$, so the expected number of colourings for triangles $D$ and $E$ would be $\frac13*2 + \frac23*1 = \frac43$. Then the expected value for the number of colourings for the whole diagram would be $$6*2^9 * \frac43 = 2^{12}.$$ But in fact the colourings of $x$ and $y$ are not independent. So the above argument is not rigorous, and the answer may not even be correct (though it must be a good approximation!).

Opalg said:
[TIKZ]\coordinate (A) at (30:2) ;
\coordinate (B) at (90:2) ;
\coordinate (C) at (150:2) ;
\coordinate (D) at (210:2) ;
\coordinate (E) at (270:2) ;
\coordinate (F) at (330:2) ;
\coordinate (U) at (0:5) ;
\coordinate (V) at (60:5) ;
\coordinate (W) at (120:5) ;
\coordinate (X) at (180:5) ;
\coordinate (Y) at (240:5) ;
\coordinate (Z) at (300:5) ;
\foreach \point in {A,B,C,D,E,F,U,V,W,X,Y,Z} \fill [black] (\point) circle (3pt) ;
\draw (A) -- (B) -- (C) -- (D) -- (E) -- (F) -- (A) -- (U) -- (V) -- (W) -- (X) -- (Y) -- (Z) -- (U) --(A) -- (V) -- (B) -- (W) -- (C) -- (X) -- (D) -- (Y) -- (E) -- node[ left ]{$x$}(Z) -- node[ right ]{$z$}(F) -- node[ below ]{$y$}(U) ;
\draw (-90:3.5) node{$6$} ;
\foreach \x in {0,30,...,240} \draw (\x:3.5) node{$2$} ;
\draw (270:2.75) node{$A$} ;
\draw (240:2.75) node{$B$} ;
\draw (0:2.75) node{$C$} ;
\draw (330:2.75) node{$D$} ;
\draw (300:2.75) node{$E$} ;[/TIKZ]
Start with the triangle labelled $A$ at the bottom of the diagram. There are 6 ways to colour its three sides in different colours. Next, look at triangle $B$. One of its sides has already been coloured, and there are 2 ways to colour the remaining sides. Continuing in this way clockwise round the diagram, there are two ways to colour each of the triangles up to and including the one labelled $C$. There are now two cases to consider for the remaining triangles $D$ and $E$. If side $x$ in triangle $A$ and side $y$ in triangle $C$ have different colours then there is only one choice for the colour of side $z$ and therefore only 1 way to colour triangles $D$ and $E$. But if $x$ and $y$ have the same colour then there are two choices for the colour of side $z$, and therefore 2 ways to colour triangles $D$ and $E$.

Here's where the argument becomes probabilistic and unreliable. If the colours of $x$ and $y$ had been chosen independently at random, then the probability of them being the same would be $\frac13$, so the expected number of colourings for triangles $D$ and $E$ would be $\frac13*2 + \frac23*1 = \frac43$. Then the expected value for the number of colourings for the whole diagram would be $$6*2^9 * \frac43 = 2^{12}.$$ But in fact the colourings of $x$ and $y$ are not independent. So the above argument is not rigorous, and the answer may not even be correct (though it must be a good approximation!).
Almost right :)

I now think that the answer should be $2^{12} + 2=4098$, but I don't have a proof of that.

## 1. How many different color combinations are possible with 12 triangles?

The number of possible color combinations with 12 triangles is 4096. This can be calculated by taking 2 to the power of 12, since each triangle can be either black or white, resulting in 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 4096.

## 2. Is there a specific pattern or formula for counting color combinations in 12 triangles?

Yes, the formula for counting color combinations in 12 triangles is 2^n, where n is the number of triangles. This is because each triangle can have 2 possible colors, resulting in a total of 2^n combinations.

## 3. Can the number of color combinations change if the number of triangles is increased or decreased?

Yes, the number of color combinations will change if the number of triangles is increased or decreased. The formula for counting color combinations is 2^n, so as n increases or decreases, the number of combinations will also change accordingly.

## 4. Do the positions of the triangles affect the number of color combinations?

Yes, the positions of the triangles can affect the number of color combinations. For example, if the triangles are arranged in a specific pattern or design, the number of color combinations may be limited. However, if the triangles are randomly placed, the number of color combinations will remain the same.

## 5. How does the concept of permutations and combinations apply to counting color combinations in 12 triangles?

The concept of permutations and combinations does not directly apply to counting color combinations in 12 triangles. This is because the order of the colors does not matter in this scenario. Each triangle can only have one color, so there are no permutations involved. The formula for counting color combinations is based on combinations, where the order of the elements does not matter.

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