Difference between revisions of "2010 AMC 10B Problems/Problem 23"

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(Problem)
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The entries in a <math>3 \times 3</math> array include all the digits from 1 through 9, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?  
 
The entries in a <math>3 \times 3</math> array include all the digits from 1 through 9, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?  
  
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\{{MAA Notice}}qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60 </math>
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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>
  
 
==Solution==
 
==Solution==

Revision as of 15:01, 21 August 2013

Problem

The entries in a $3 \times 3$ array include all the digits from 1 through 9, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$

Solution

By the hook-length formula, the answer is $\frac{9!}{5\cdot 4^{2}\cdot 3^{3}\cdot 2^{2}\cdot 1}= \boxed{\textbf{(D)}\ 42}$

See also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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 10 Problems and Solutions

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