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

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==Solution==
 
==Solution==
 
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By the [http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation hook-length formula], the answer is <math> \frac{9!}{5\cdot 4^{2}\cdot 3^{3}\cdot 2^{2}\cdot 1}= \boxed{\textbf{(D)}\ 42}</math>
By the [http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation hook-length formula], the answer is <math> \frac{9!}{5\cdot 4^{2}\cdot 3^{3}\cdot 2^{2}\cdot 1}= 42\ \textbf{(D)} </math>
 
 
 
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2010|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2010|ab=B|num-b=22|num-a=24}}

Revision as of 15:41, 26 November 2011

Problem

The entries in a $3x3$ 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