Difference between revisions of "2013 AIME II Problems/Problem 9"

(Solution 1.1)
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We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
 
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
  
* Three pieces: <math>5+1+1</math>, <math>4+2+1</math>, <math>4+1+2</math>, etc, <math>\dbinom{6}{2}=15</math> ways in total
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* Three pieces: <math>5+1+1</math>, <math>4+2+1</math>, <math>4+1+2</math>, etc, <math>\dbinom{6}{2}=15</math> ways in total (just apply stars and bars here)
 
* Four pieces: <math>\dbinom{6}{3}=20</math>
 
* Four pieces: <math>\dbinom{6}{3}=20</math>
 
* Five pieces: <math>\dbinom{6}{4}=15</math>
 
* Five pieces: <math>\dbinom{6}{4}=15</math>
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<math>\sum_{i=0}^{7} (</math>the amount of ways to divide the board into <math>i</math> pieces<math>) \cdot (</math>the amount of ways to color the respective divisions)
 
<math>\sum_{i=0}^{7} (</math>the amount of ways to divide the board into <math>i</math> pieces<math>) \cdot (</math>the amount of ways to color the respective divisions)
  
The amount of ways to divide the board is just stars and bars. The colorings are PIE giving <math>3^i-3\cdot2^i+3</math>. Plus, you don't have to worry about <math>i\ge2</math> since they all give no solutions. So our equation becomes:
+
The amount of ways to divide the board is just stars and bars. The colorings are PIE giving <math>3^i-3\cdot2^i+3</math>. Plus, you don't have to worry about the cases where i=1 or i=2 since they both give no solutions. So our equation becomes:
  
 
<math>\sum_{i=3}^{7} (\dbinom{6}{i})\cdot(3^i-3\cdot2^i+3)</math>
 
<math>\sum_{i=3}^{7} (\dbinom{6}{i})\cdot(3^i-3\cdot2^i+3)</math>
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~Rowechen Zhong
 
~Rowechen Zhong
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==Solution 2 (Recursive)==
 
==Solution 2 (Recursive)==
 
3 colors is a lot. How many ways can we tile an n x 1 board with one color? It's going to be <math>2^{n-1}</math> because if you draw out the <math>n</math> spaces, you can decide whether each of the borders between the tiles are either there or not there. There are <math>n-1</math> borders so there are <math>2^{n-1}</math> tilings. Define a one-tiling of an mx1 as <math>f_1(m)</math>
 
3 colors is a lot. How many ways can we tile an n x 1 board with one color? It's going to be <math>2^{n-1}</math> because if you draw out the <math>n</math> spaces, you can decide whether each of the borders between the tiles are either there or not there. There are <math>n-1</math> borders so there are <math>2^{n-1}</math> tilings. Define a one-tiling of an mx1 as <math>f_1(m)</math>
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Finally, we just sorta bash through the computation to get <math>f_3(7)=8\boxed{106}</math>
 
Finally, we just sorta bash through the computation to get <math>f_3(7)=8\boxed{106}</math>
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 +
==Solution 3==
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Let <math>n</math> be the length of the board and <math>x</math> be the number of colors. We will find the number of ways to tile the <math>n \times 1</math> board with no color restrictions (some colors may be unused) and then use PIE.
 +
 +
By stars and bars, the number of ways to divide the board into <math>k</math> pieces is <math>{n-1 \choose k-1}</math>. There are <math>x^k</math> ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is
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<cmath>\begin{align*}
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\chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\
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&= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\
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&= x(x+1)^{n-1}.
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\end{align*}</cmath>
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 +
In the given problem, we have <math>n=7</math>. By PIE, we have
 +
<cmath>\begin{align*}
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&\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\
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&= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\
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&= 8106 \rightarrow \boxed{106}.
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\end{align*}</cmath>
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 +
The above formula for <math>\chi(n, x)</math> can also be derived directly as follows:
 +
 +
* The first square can be colored in one of <math>x</math> ways.
 +
* Proceeding left to right, at each of the remaining <math>n-1</math> squares, we have <math>x+1</math> options:
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** <math>x</math> ways to begin a new tile, selecting an arbitrary color. Note that this includes the case where we begin a new tile of the same color.
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** <math>1</math> way to expand the current tile by attaching a square of the current color.
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 +
This results in <math>x(x+1)^{n-1}</math> overall ways to divide and color the board (without the color use restriction).
  
 
==See Also==
 
==See Also==

Latest revision as of 09:45, 27 October 2020

Problem 9

A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$.

Solution 1

Firstly, we consider how many different ways possible to divide the $7\times 1$ board. We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.

  • Three pieces: $5+1+1$, $4+2+1$, $4+1+2$, etc, $\dbinom{6}{2}=15$ ways in total (just apply stars and bars here)
  • Four pieces: $\dbinom{6}{3}=20$
  • Five pieces: $\dbinom{6}{4}=15$
  • Six pieces: $\dbinom{6}{5}=6$
  • Seven pieces: $\dbinom{6}{6}=1$

Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:

  • Three pieces: $3^3-3\times 2^3+3=6$
  • Four pieces: $3^4-3\times 2^4+3=36$
  • Five pieces: $3^5-3\times 2^5+3=150$
  • Six pieces: $3^6-3\times 2^6+3=540$
  • Seven pieces: $3^7-3\times 2^7+3=1806$

Finally, we combine them together: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$.

So the answer is $\boxed{106}$.

Solution 1.1 (Faster/Streamlined 1)

This solution is basically solution 1 with more things done at once. The game plan:

$\sum_{i=0}^{7} ($the amount of ways to divide the board into $i$ pieces$) \cdot ($the amount of ways to color the respective divisions)

The amount of ways to divide the board is just stars and bars. The colorings are PIE giving $3^i-3\cdot2^i+3$. Plus, you don't have to worry about the cases where i=1 or i=2 since they both give no solutions. So our equation becomes:

$\sum_{i=3}^{7} (\dbinom{6}{i})\cdot(3^i-3\cdot2^i+3)$

Write it all out, keep the numbers small with mod 1000, and you'll eventually arrive at your answer of $\boxed{106}$.

~Rowechen Zhong

Solution 2 (Recursive)

3 colors is a lot. How many ways can we tile an n x 1 board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$

Now let's look at two colors. Let's see if we can get a two-tiling of an (n+1) x 1 based off a n x 1. There are 2 cases we should consider:

1) The n x 1 was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$

2) The n x 1 was a two-color tiling so now we've got 3 cases to form the (n+1) x 1: We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the n x 1.

This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$

Time to tackle the 3-color tiling. Again, we split into 2 cases:

1) The n x 1 was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by 3C2 because we have to pick 2 different colors for the two-color tiling.

2) The n x 1 was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the n x 1. This gives $4f_3(n)$

So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$.

Finally, we just sorta bash through the computation to get $f_3(7)=8\boxed{106}$

Solution 3

Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \times 1$ board with no color restrictions (some colors may be unused) and then use PIE.

By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \choose k-1}$. There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \begin{align*} \chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\ &= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\ &= x(x+1)^{n-1}. \end{align*}

In the given problem, we have $n=7$. By PIE, we have \begin{align*} &\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\ &= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\ &= 8106 \rightarrow \boxed{106}. \end{align*}

The above formula for $\chi(n, x)$ can also be derived directly as follows:

  • The first square can be colored in one of $x$ ways.
  • Proceeding left to right, at each of the remaining $n-1$ squares, we have $x+1$ options:
    • $x$ ways to begin a new tile, selecting an arbitrary color. Note that this includes the case where we begin a new tile of the same color.
    • $1$ way to expand the current tile by attaching a square of the current color.

This results in $x(x+1)^{n-1}$ overall ways to divide and color the board (without the color use restriction).

See Also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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