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Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 5"

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== Solution ==
 
== Solution ==
 
(a) Let us write the position of any coin as <math>(r,c)</math>. Then, <math>0 \le r \le 3, 0 \le c \le 3</math>. Therefore, there are <math>4 \times 4 = 16</math> different positions for the first coin, <math>3 \times 3 = 9</math> possibilities for the second coin, <math>2 \times 2 = 4</math> possibilities for the third coin, and <math>1 \times 1 = 1</math> possibility for the fourth coin. Together, there are <math>16 \times 9 \times 4 \times 1 = 576</math> possible placements of the two coins.
 
(a) Let us write the position of any coin as <math>(r,c)</math>. Then, <math>0 \le r \le 3, 0 \le c \le 3</math>. Therefore, there are <math>4 \times 4 = 16</math> different positions for the first coin, <math>3 \times 3 = 9</math> possibilities for the second coin, <math>2 \times 2 = 4</math> possibilities for the third coin, and <math>1 \times 1 = 1</math> possibility for the fourth coin. Together, there are <math>16 \times 9 \times 4 \times 1 = 576</math> possible placements of the two coins.
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 +
(b)Using the same process above, the possible positions of a coin at <math>(r,c)</math>, where  <math>0 \le r \le N-1, 0 \le c \le N-1</math>, can be counted as <math>(N-1)^2</math> possibilities for the first coin, <math>(N-2)^2</math> possibilities for the second coin, <math>(N-3)^2</math> possibilities for the third coin, and <math>(N-4)^2</math> possibilities for the fourth coin. Finally, there are <math>(N-1)^2 + (N-2)^2 + (N-3)^2 + (N-4)^2 = 4N^2 -20N + 30</math> possibilities.
  
 
== See also ==
 
== See also ==

Revision as of 20:55, 9 February 2015

Problem

(a) In the $4 \times 4$ grid shown, four coins are randomly placed in different squares. What is the probability that no two coins lie in the same row or column?

$\begin{tabular}{|c|c|c|c|} \hline &&& \\ \hline &&& \\ \hline &&& \\ \hline &&& \\ \hline \end{tabular}$

(b) Generalize this to an $N \times N$ grid.


Solution

(a) Let us write the position of any coin as $(r,c)$. Then, $0 \le r \le 3, 0 \le c \le 3$. Therefore, there are $4 \times 4 = 16$ different positions for the first coin, $3 \times 3 = 9$ possibilities for the second coin, $2 \times 2 = 4$ possibilities for the third coin, and $1 \times 1 = 1$ possibility for the fourth coin. Together, there are $16 \times 9 \times 4 \times 1 = 576$ possible placements of the two coins.

(b)Using the same process above, the possible positions of a coin at $(r,c)$, where $0 \le r \le N-1, 0 \le c \le N-1$, can be counted as $(N-1)^2$ possibilities for the first coin, $(N-2)^2$ possibilities for the second coin, $(N-3)^2$ possibilities for the third coin, and $(N-4)^2$ possibilities for the fourth coin. Finally, there are $(N-1)^2 + (N-2)^2 + (N-3)^2 + (N-4)^2 = 4N^2 -20N + 30$ possibilities.

See also

2010 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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