Difference between revisions of "2005 Alabama ARML TST Problems/Problem 5"
m |
(→Solution) |
||
| (8 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | A <math> | + | A <math>2\times 2</math> square grid is constructed with four <math>1\times 1</math> squares. The square on the upper left is labeled <i>A</i>, the square on the upper right is labeled <i>B</i>, the square in the lower left is labeled <i>C</i>, and the square on the lower right is labeled <i>D</i>. The four squares are to be painted such that 2 are blue, 1 is red, and 1 is green. In how many ways can this be done? |
| + | |||
==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
We choose two squares to be blue and one to be red; then the green's position is forced. There are <math>{4 \choose 2}{2 \choose 1}=12</math> ways to do this. | We choose two squares to be blue and one to be red; then the green's position is forced. There are <math>{4 \choose 2}{2 \choose 1}=12</math> ways to do this. | ||
| − | {{ | + | Equivalently, we could choose one square to be red and one square to be green, then blue is forced: <math>\binom{4}{1} \binom{3}{1}=12</math>. |
| + | |||
| + | ===Solution 2=== | ||
| + | Since rotations and reflections are not distinct, the number of ways is just <math>\frac{4!}{2! \cdot 1! \cdot 1!} = 12</math> | ||
| + | |||
==See Also== | ==See Also== | ||
| − | + | {{ARML box|year=2005|state=Alabama|num-b=4|num-a=6}} | |
| − | + | ||
| − | + | [[Category:Intermediate Combinatorics Problems]] | |
Latest revision as of 01:36, 3 January 2023
Problem
A 
square grid is constructed with four 
squares. The square on the upper left is labeled A, the square on the upper right is labeled B, the square in the lower left is labeled C, and the square on the lower right is labeled D. The four squares are to be painted such that 2 are blue, 1 is red, and 1 is green. In how many ways can this be done?
Solution
Solution 1
We choose two squares to be blue and one to be red; then the green's position is forced. There are 
ways to do this.
Equivalently, we could choose one square to be red and one square to be green, then blue is forced: 
.
Solution 2
Since rotations and reflections are not distinct, the number of ways is just 
See Also
| 2005 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 4 |
Followed by: Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
