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, | + | (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. |
== See also == | == See also == | ||
Revision as of 20:52, 9 February 2015
Problem
(a) In the 
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?
(b) Generalize this to an 
grid.
Solution
(a) Let us write the position of any coin as 
. Then, 
. Therefore, there are 
different positions for the first coin, 
possibilities for the second coin, 
possibilities for the third coin, and 
possibility for the fourth coin. Together, there are 
possible placements of the two coins.
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 | ||
