Difference between revisions of "2020 AMC 10B Problems/Problem 5"

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<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>
  
==Solution==
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==Solution 1==
  
 
Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
 
Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
  
There are <math>7!</math> ways to order <math>7</math> objects. However, since there's <math>3!=6</math> ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and <math>2!=2</math> ways to order the green tiles, we have to divide out these possibilities.
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There are <math>7!</math> ways to order <math>7</math> objects. However, since there's <math>3!=6</math> ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and <math>2!=2</math> ways to order the green tiles, we have to divide out these possibilities: <math>\frac{7!}{3! \cdot2}=\boxed{\textbf{(B) }420}</math>.  
  
<math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88
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~quacker88
  
==Solution 2 (2 second solve)==
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==Solution 2==
  
Be a memer and guess <math>\boxed{\textbf{(B) }420}</math>, which happens to be the correct answer.
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There are <math>7</math> choose <math>3</math> ways to arrange the yellow tiles which is <math>35</math>.
-fidgetboss_4000
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Then from the remaining tiles there are <math>\binom{4}{2}=6</math> ways to arrange the green tiles.
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And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles,
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giving us an answer of <math>35\cdot6\cdot2=420</math>.
  
==Video Solution==
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~noahdavid (Edited by starshooter11)
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==Solution 3==
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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}.</math>
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~@azure123456 BZ
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== Video Solutions ==
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===Video Solution 1===
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https://youtu.be/0W3VmFp55cM?t=540
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~ pi_is_3.14
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===Video Solution 2===
 
https://youtu.be/Gkm5rU5MlOU
 
https://youtu.be/Gkm5rU5MlOU
  
 
~IceMatrix
 
~IceMatrix
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===Video Solution 3===
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https://youtu.be/4vz8IPfzs2c
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 18:29, 6 September 2021

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.)

$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\  630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

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 $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to order the green tiles, we have to divide out these possibilities: $\frac{7!}{3! \cdot2}=\boxed{\textbf{(B) }420}$.

~quacker88

Solution 2

There are $7$ choose $3$ ways to arrange the yellow tiles which is $35$. Then from the remaining tiles there are $\binom{4}{2}=6$ 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 $35\cdot6\cdot2=420$.

~noahdavid (Edited by starshooter11)

Solution 3

We can choose a different frame to solve this problem. Our tile combination can be written as $Y, Y, Y, G, G, B, P.$ We can focus on $Y, Y, Y, G, G$ first, which gives us $\binom{5}{3}=10.$ Now we can insert our brown tile into this which only has $6$ choices(like $Y,B, Y, Y, G, G$ and $Y, Y, Y, G, G, B$ etc.), then insert purple tile which only has $7$ choices(like $B,Y, P, Y, Y, G, G$ and $B, Y, Y, Y, G, G, P$ etc.). Multiply them together we get $10\cdot 6\cdot 7=\boxed{420}.$

~@azure123456 BZ

Video Solutions

Video Solution 1

https://youtu.be/0W3VmFp55cM?t=540

~ pi_is_3.14

Video Solution 2

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

Video Solution 3

https://youtu.be/4vz8IPfzs2c

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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

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