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Difference between revisions of "2001 Pan African MO Problems/Problem 2"

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Summing all possible values of <math>k</math> means that there are a total of <math>\binom{n-1}{n-1} + \binom{n-1}{n-2} + \cdots + \binom{n-1}{0} = \boxed{2^{n-1}}</math> walls with <math>n</math> cubes.
 
Summing all possible values of <math>k</math> means that there are a total of <math>\binom{n-1}{n-1} + \binom{n-1}{n-2} + \cdots + \binom{n-1}{0} = \boxed{2^{n-1}}</math> walls with <math>n</math> cubes.
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==Solution 2==
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There are <math>n</math> cubes. After placing the first cube down, we have <math>(n-1)</math> cubes left. Now for each of these remaining <math>(n-1)</math> remaining cubes, we have two options; stack the cube or put it in front. This then gives that since there are <math>2</math> options for each of the <math>(n-1)</math> cubes, the answer as <math>\boxed{2^{n-1}}</math>.
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-th1nq3r
  
 
==See Also==
 
==See Also==

Latest revision as of 07:39, 5 September 2021

Problem

Let $n$ be a positive integer. A child builds a wall along a line with $n$ identical cubes. He lays the first cube on the line and at each subsequent step, he lays the next cube either on the ground or on the top of another cube, so that it has a common face with the previous one. How many such distinct walls exist?

Solution

From smaller values of $n$, there is 1 wall with 1 cube, 2 walls with 2 cubes, 4 walls with 3 cubes, and 8 walls with 4 cubes. Thus, we can suspect that there are $2^{n-1}$ walls with $n$ cubes.


To prove our claim, we can calculate the number of walls with $n$ blocks and $k$ columns. We can use ball-and-urn counting to determine the number of walls. Since there are $k$ columns, there would be $k-1$ dividers. There are a total of $n$ blocks, but each column must have at least one block, so there are $n-k$ blocks left to sort. Thus, there are $\binom{n-1}{n-k}$ walls that have $n$ blocks and $k$ columns.


Summing all possible values of $k$ means that there are a total of $\binom{n-1}{n-1} + \binom{n-1}{n-2} + \cdots + \binom{n-1}{0} = \boxed{2^{n-1}}$ walls with $n$ cubes.

Solution 2

There are $n$ cubes. After placing the first cube down, we have $(n-1)$ cubes left. Now for each of these remaining $(n-1)$ remaining cubes, we have two options; stack the cube or put it in front. This then gives that since there are $2$ options for each of the $(n-1)$ cubes, the answer as $\boxed{2^{n-1}}$.

-th1nq3r

See Also

2001 Pan African MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All Pan African MO Problems and Solutions
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