Difference between revisions of "2020 AMC 10B Problems/Problem 5"
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==Problem== | ==Problem== | ||
| − | How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.) | + | How many distinguishable arrangements are there of <math>1</math> brown tile, <math>1</math> purple tile, <math>2</math> green tiles, and <math>3</math> yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.) |
<math>\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050</math> | <math>\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050</math> | ||
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==Solution 3== | ==Solution 3== | ||
| − | We can choose a different frame to solve this problem. Our tile combination can be written as <math>Y, Y, Y, G, G, B, P.</math> We can focus on <math>Y, Y, Y, G, G</math> first, which gives us <math>\binom{5}{3}=10.</math> Now we can insert our brown tile into this which only has <math>6</math> choices(like <math>Y,B, Y, Y, G, G</math> and <math>Y, Y, Y, G, G, B</math> etc.), then insert purple tile which only has <math>7</math> choices(like <math>B,Y, P, Y, Y, G, G</math> and <math>B, Y, Y, Y, G, G, P</math> etc.). Multiply them together we get <math>10\cdot 6\cdot 7=\boxed{420} | + | We can choose a different frame to solve this problem. Our tile combination can be written as <math>Y, Y, Y, G, G, B, P.</math> We can focus on <math>Y, Y, Y, G, G</math> first, which gives us <math>\binom{5}{3}=10.</math> Now we can insert our brown tile into this which only has <math>6</math> choices(like <math>Y,B, Y, Y, G, G</math> and <math>Y, Y, Y, G, G, B</math> etc.), then insert purple tile which only has <math>7</math> choices(like <math>B,Y, P, Y, Y, G, G</math> and <math>B, Y, Y, Y, G, G, P</math> etc.). Multiply them together we get <math>10\cdot 6\cdot 7=\boxed{\textbf{(B)}\ 420}</math>. |
~@azure123456 BZ | ~@azure123456 BZ | ||
| − | == Video | + | == Solution 4 (Concise)== |
| − | ===Video Solution | + | |
| + | Let <math>B</math> be brown, <math>P</math> be purple, <math>G</math> be green, and <math>Y</math> be yellow. Then, we are just ordering <math>Y</math>, <math>Y</math>, <math>Y</math>, <math>G</math>, <math>G</math>, <math>B</math>, and <math>P</math>. Hence, <math>\frac{7!}{3! \cdot 2!} = \boxed{\textbf{(B)}\ 420}</math>. | ||
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| + | ~MrThinker | ||
| + | |||
| + | ==Video Solution (HOW TO CRITICALLY THINK!!!)== | ||
| + | https://youtu.be/SQ8XRpl2gXE | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | ===Video Solution == | ||
https://youtu.be/0W3VmFp55cM?t=540 | https://youtu.be/0W3VmFp55cM?t=540 | ||
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===Video Solution 2=== | ===Video Solution 2=== | ||
https://youtu.be/Gkm5rU5MlOU | https://youtu.be/Gkm5rU5MlOU | ||
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===Video Solution 3=== | ===Video Solution 3=== | ||
https://youtu.be/4vz8IPfzs2c | https://youtu.be/4vz8IPfzs2c | ||
Latest revision as of 13:55, 6 June 2023
Contents
Problem
How many distinguishable arrangements are there of 
brown tile, purple tile, 
green tiles, and 
yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
Solution 1
Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
There are 
ways to order 
objects. However, since there's 
ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and 
ways to order the green tiles, we have to divide out these possibilities: 
.
~quacker88
Solution 2
There are choose ways to arrange the yellow tiles which is 
.
Then from the remaining tiles there are 
ways to arrange the green tiles.
And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles,
giving us an answer of 
.
~noahdavid (Edited by starshooter11)
Solution 3
We can choose a different frame to solve this problem. Our tile combination can be written as 
We can focus on 
first, which gives us 
Now we can insert our brown tile into this which only has 
choices(like 
and 
etc.), then insert purple tile which only has choices(like 
and 
etc.). Multiply them together we get 
.
~@azure123456 BZ
Solution 4 (Concise)
Let 
be brown, 
be purple, 
be green, and 
be yellow. Then, we are just ordering , , , , , , and . Hence, 
.
~MrThinker
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
=Video Solution
https://youtu.be/0W3VmFp55cM?t=540
Video Solution 2
Video Solution 3
~savannahsolver
Video Solution 4
https://youtu.be/waxVSOt_v1M?t=275
~ AMBRIGGS
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
