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Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 10"

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== Solution ==
 
== Solution ==
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(a) <math>14</math> (b) <math>132</math>
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(c) <math>\frac{1}{n + 1} \binom{2n}{n}</math> <math>= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1}  \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}</math> <math>=</math> nth [[Catalan number]]
  
 
== See also ==
 
== See also ==
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{{UNCO Math Contest box|year=2015|n=II|num-b=9|after=BONUS}}
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{{UNCO Math Contest box|year=2015|n=II|num-b=9|after=[[2015 UNCO Math Contest II Problems/BONUS| BONUS]]}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 18:54, 19 December 2023

Problem

$\begin{tabular}[t]{|c|c|c|c|}\hline & & & \\\hline & & & \\\hline \end{tabular}$

(a) You want to arrange $8$ biologists of $8$ different heights in two rows for a photograph. Each row must have $4$ biologists. Height must increase from left to right in each row. Each person in back must be taller than the person directly in front of him. How many different arrangements are possible?


$\begin{tabular}[t]{|c|c|c|c|c|c|}\hline & & & & & \\\hline & & & & & \\\hline \end{tabular}$

(b) You arrange $12$ biologists of $12$ different heights in two rows of $6$, with the same conditions on height as in part (a). How many different arrangements are possible? Remember to justify your answers.


(c) You arrange $2n$ biologists of $2n$ different heights in two rows of $n$, with the same conditions on height as in part (a). Give a formula in terms of $n$ for the number of possible arrangements.

Solution

(a) $14$ (b) $132$

(c) $\frac{1}{n + 1} \binom{2n}{n}$ $= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}$ $=$ nth Catalan number

See also

2015 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
BONUS
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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