Difference between revisions of "2011 AMC 10A Problems/Problem 22"
(→Solution 3) |
|||
Line 31: | Line 31: | ||
==Solution 3== | ==Solution 3== | ||
− | + | There are <math>6</math> ways to assign a color to <math>A</math>. WLOG, give vertex <math>A</math> a color; we can multiply by <math>6</math> at the end. Since vertices <math>A</math> and <math>C</math> cannot have the same color, there are <math>5</math> ways to assign colors to vertex <math>C</math>. Using this same logic, there are <math>5</math> ways to assign a color to vertices <math>E</math>, <math>B</math>, and <math>D</math>, giving a total of <math>5^4=625</math> ways. However, if vertex <math>D</math> cannot be the same color as vertex <math>A</math>. To use complementary counting, we need to find the amount of ways for <math>D</math> and <math>A</math> to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are <math>5^3=125</math> ways, except we must subtract the number of ways for a triangle. Each time, there is <math>1</math> less vertex, so <math>5</math> times less ways to color. This process stops when there are only <math>2</math> vertices left; in this case there are simply <math>5</math> ways to color this figure. | |
− | + | So in conclusion, there are <math>6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120 \ \mathbf{(C)}}</math> ways. | |
== See Also == | == See Also == |
Revision as of 17:00, 11 September 2018
Contents
[hide]Problem 22
Each vertex of convex pentagon is to be assigned a color. There are
colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution 1
Let vertex be any vertex, then vertex
be one of the diagonal vertices to
,
be one of the diagonal vertices to
, and so on. We consider cases for this problem.
In the case that has the same color as
,
has a different color from
and so
has a different color from
and
. In this case,
has
choices,
has
choices (any color but the color of
),
has
choice,
has
choices, and
has
choices, resulting in a possible of
combinations.
In the case that has a different color from
and
has a different color from
,
has
choices,
has
choices,
has
choices (since
and
necessarily have different colors),
has
choices, and
has
choices, resulting in a possible of
combinations.
In the case that has a different color from
and
has the same color as
,
has
choices,
has
choices,
has
choices,
has
choice, and
has
choices, resulting in a possible of
combinations.
Adding all those combinations up, we get .
Solution 2
First, notice that there can be cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case : There are
ways of assigning each vertex a different color.
Case : There are
ways. After picking four colors, we can rotate our pentagon in
ways to get different outcomes.
Case : There are
ways of arranging the final case. We can pick
colors for our pentagon. There are
spots for the first pair of colors. Then, there are
possible ways we can put the final pair in the last
spaces. But because the two pairs are indistinguishable, we divide by
.
Adding all the possibilities up, we get
~ZericHang
Solution 3
There are ways to assign a color to
. WLOG, give vertex
a color; we can multiply by
at the end. Since vertices
and
cannot have the same color, there are
ways to assign colors to vertex
. Using this same logic, there are
ways to assign a color to vertices
,
, and
, giving a total of
ways. However, if vertex
cannot be the same color as vertex
. To use complementary counting, we need to find the amount of ways for
and
to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are
ways, except we must subtract the number of ways for a triangle. Each time, there is
less vertex, so
times less ways to color. This process stops when there are only
vertices left; in this case there are simply
ways to color this figure.
So in conclusion, there are
ways.
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.