Difference between revisions of "2010 AMC 12B Problems/Problem 17"
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<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math> | <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math> | ||
| − | == Solution == | + | == Solution 1== |
The first 4 numbers will form one of 3 tetris "shapes". | The first 4 numbers will form one of 3 tetris "shapes". | ||
| Line 72: | Line 72: | ||
The answer is <math> 4\times 6 + 2\times 9 = \boxed{\text{(D) }42}</math>. | The answer is <math> 4\times 6 + 2\times 9 = \boxed{\text{(D) }42}</math>. | ||
| + | |||
| + | |||
| + | == Solution 2== | ||
| + | This solution is trivial by the hook length theorem. The hooks look like this: | ||
| + | |||
| + | <math> \begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ | ||
| + | \hline 4 & 3 & 2\\ | ||
| + | \hline 3 & 2 & 1\\ | ||
| + | \hline \end{tabular}</math> | ||
| + | |||
| + | So, the answer is <math>\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}</math> = <math>\boxed{\text{(D) }42}</math> | ||
== See also == | == See also == | ||
Revision as of 13:25, 1 July 2015
Contents
Problem
The entries in a 
array include all the digits from 
through 
, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
Solution 1
The first 4 numbers will form one of 3 tetris "shapes".
First, let's look at the numbers that form a 2x2 block, sometimes called tetris 
:
Second, let's look at the numbers that form a vertical "L", sometimes called tetris 
:
Third, let's look at the numbers that form a horizontal "L", sometimes called tetris 
:
Now, the numbers 6-9 will form similar shapes (rotated by 180 degrees, and anchored in the lower-right corner of the 3x3 grid).
If you match up one tetris shape from the numbers 1-4 and one tetris shape from the numbers 6-9, there is only one place left for the number 5 to be placed.
So what shapes will physically fit in the 3x3 grid, together?
The answer is 
.
Solution 2
This solution is trivial by the hook length theorem. The hooks look like this:
So, the answer is 
= 
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
