https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Gsaelite&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:20:23ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_16&diff=900842011 AMC 12A Problems/Problem 162018-01-30T16:51:27Z<p>Gsaelite: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math><br />
\textbf{(A)}\ 2520 \qquad<br />
\textbf{(B)}\ 2880 \qquad<br />
\textbf{(C)}\ 3120 \qquad<br />
\textbf{(D)}\ 3250 \qquad<br />
\textbf{(E)}\ 3750 </math><br />
<br />
== Solution ==<br />
<br />
We can do some casework when working our way around the pentagon from <math>A</math> to <math>E</math>. At each stage, there will be a makeshift diagram.<br />
<br />
1.) For <math>A</math>, we can choose any of the 6 colors.<br />
<br />
A : 6<br />
<br />
2.) For <math>B</math>, we can either have the same color as <math>A</math>, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and <math>D</math> will be affected by both <math>A</math> and <math>B</math>.<br />
<br />
A : 6<br />
B:1 B:5<br />
<br />
3.) For <math>C</math>, we cannot have the same color as <math>A</math>. Also, we can have the same color as <math>B</math> (<math>E</math> will be affected), or any of the other 4 colors. Because <math>C</math> can't be the same as <math>A</math>, it can't be the same as <math>B</math> if <math>B</math> is the same as <math>A</math>, so it can be any of the 5 other colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
<br />
4.) <math>D</math> is affected by <math>A</math> and <math>B</math>. If they are the same, then <math>D</math> can be any of the other 5 colors. If they are different, then <math>D</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
<br />
5.) <math>E</math> is affected by <math>B</math> and <math>C</math>. If they are the same, then <math>E</math> can be any of the other 5 colors. If they are different, then <math>E</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
E:4 E:4 E:5<br />
<br />
6.) Now, we can multiply these three paths and add them:<br />
<math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math><br />
<br />
7.) Our answer is <math>C</math>!<br />
<br />
==Solution 2==<br />
<br />
Right off the bat, we can analyze three things: <br />
<br />
<br />
1.) There can only be two of the same color on the pentagon.<br />
<br />
2.) Any pair of the same color can only be next to each other on the pentagon.<br />
<br />
3.) There can only be two different pairs of same colors on the pentagon at once.<br />
<br />
<br />
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.<br />
<br />
<br />
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count <math>6!=720</math> cases. No rotation is necessary because all permutations are accounted for.<br />
<br />
<br />
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex. <br />
<br />
We get <math>6\times5\times4\times3=360</math>.<br />
<br />
However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for one pair.<br />
<br />
<br />
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.<br />
<br />
We get <math>6\times5\times4=120</math>.<br />
<br />
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math> possibilities for two pairs. <br />
<br />
<br />
5.) If we add all of three cases together, we get <math>720+1800+600=3120</math>. The answer is <math>C</math>.<br />
<br />
Solution by gsaelite<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Gsaelitehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_16&diff=900832011 AMC 12A Problems/Problem 162018-01-30T16:50:24Z<p>Gsaelite: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math><br />
\textbf{(A)}\ 2520 \qquad<br />
\textbf{(B)}\ 2880 \qquad<br />
\textbf{(C)}\ 3120 \qquad<br />
\textbf{(D)}\ 3250 \qquad<br />
\textbf{(E)}\ 3750 </math><br />
<br />
== Solution ==<br />
<br />
We can do some casework when working our way around the pentagon from <math>A</math> to <math>E</math>. At each stage, there will be a makeshift diagram.<br />
<br />
1.) For <math>A</math>, we can choose any of the 6 colors.<br />
<br />
A : 6<br />
<br />
2.) For <math>B</math>, we can either have the same color as <math>A</math>, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and <math>D</math> will be affected by both <math>A</math> and <math>B</math>.<br />
<br />
A : 6<br />
B:1 B:5<br />
<br />
3.) For <math>C</math>, we cannot have the same color as <math>A</math>. Also, we can have the same color as <math>B</math> (<math>E</math> will be affected), or any of the other 4 colors. Because <math>C</math> can't be the same as <math>A</math>, it can't be the same as <math>B</math> if <math>B</math> is the same as <math>A</math>, so it can be any of the 5 other colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
<br />
4.) <math>D</math> is affected by <math>A</math> and <math>B</math>. If they are the same, then <math>D</math> can be any of the other 5 colors. If they are different, then <math>D</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
<br />
5.) <math>E</math> is affected by <math>B</math> and <math>C</math>. If they are the same, then <math>E</math> can be any of the other 5 colors. If they are different, then <math>E</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
E:4 E:4 E:5<br />
<br />
6.) Now, we can multiply these three paths and add them:<br />
<math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math><br />
<br />
7.) Our answer is <math>C</math>!<br />
<br />
==Solution 2==<br />
<br />
Right off the bat, we can analyze three things: <br />
<br />
<br />
1.) There can only be two of the same color on the pentagon.<br />
<br />
2.) Any pair of the same color can only be next to each other on the pentagon.<br />
<br />
3.) There can only be two different pairs of same colors on the pentagon at once.<br />
<br />
<br />
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.<br />
<br />
<br />
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count <math>6!=720</math> cases. No rotation is necessary because all permutations are accounted for.<br />
<br />
<br />
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex. <br />
<br />
We get <math>6\times5\times4\times3=360</math>.<br />
<br />
However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for two pairs.<br />
<br />
<br />
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.<br />
<br />
We get <math>6\times5\times4=120</math>.<br />
<br />
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math>. <br />
<br />
<br />
5.) If we add all of three cases together, we get <math>720+1800+600=3120</math>. The answer is <math>C</math>.<br />
<br />
Solution by gsaelite<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Gsaelitehttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_16&diff=900822011 AMC 12A Problems/Problem 162018-01-30T16:49:34Z<p>Gsaelite: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br />
<br />
<math><br />
\textbf{(A)}\ 2520 \qquad<br />
\textbf{(B)}\ 2880 \qquad<br />
\textbf{(C)}\ 3120 \qquad<br />
\textbf{(D)}\ 3250 \qquad<br />
\textbf{(E)}\ 3750 </math><br />
<br />
== Solution ==<br />
<br />
We can do some casework when working our way around the pentagon from <math>A</math> to <math>E</math>. At each stage, there will be a makeshift diagram.<br />
<br />
1.) For <math>A</math>, we can choose any of the 6 colors.<br />
<br />
A : 6<br />
<br />
2.) For <math>B</math>, we can either have the same color as <math>A</math>, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and <math>D</math> will be affected by both <math>A</math> and <math>B</math>.<br />
<br />
A : 6<br />
B:1 B:5<br />
<br />
3.) For <math>C</math>, we cannot have the same color as <math>A</math>. Also, we can have the same color as <math>B</math> (<math>E</math> will be affected), or any of the other 4 colors. Because <math>C</math> can't be the same as <math>A</math>, it can't be the same as <math>B</math> if <math>B</math> is the same as <math>A</math>, so it can be any of the 5 other colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
<br />
4.) <math>D</math> is affected by <math>A</math> and <math>B</math>. If they are the same, then <math>D</math> can be any of the other 5 colors. If they are different, then <math>D</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
<br />
5.) <math>E</math> is affected by <math>B</math> and <math>C</math>. If they are the same, then <math>E</math> can be any of the other 5 colors. If they are different, then <math>E</math> can be any of the (6-2)=4 colors.<br />
<br />
A : 6<br />
B:1 B:5<br />
C:5 C:4 C:1<br />
D:5 D:4 D:4<br />
E:4 E:4 E:5<br />
<br />
6.) Now, we can multiply these three paths and add them:<br />
<math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math><br />
<br />
7.) Our answer is <math>C</math>!<br />
<br />
==Solution 2==<br />
<br />
Right off the bat, we can analyze three things: <br />
<br />
<br />
1.) There can only be two of the same color on the pentagon.<br />
<br />
2.) Any pair of the same color can only be next to each other on the pentagon.<br />
<br />
3.) There can only be two different pairs of same colors on the pentagon at once.<br />
<br />
<br />
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.<br />
<br />
<br />
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count <math>6!=720</math> cases. No rotation is necessary because all permutations are accounted for.<br />
<br />
<br />
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex. <br />
<br />
We get <math>6\times5\times4\times3=360</math>.<br />
<br />
However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for two pairs.<br />
<br />
<br />
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.<br />
<br />
We get <math>6\times5\times4=120</math>.<br />
<br />
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math>. <br />
<br />
<br />
5.) If we add all of three cases together, we get <math>720+1800+600=3120</math>. The answer is <math>C</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Gsaelite