Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Video Solutions)
 
(13 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==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>
Line 9: Line 9:
 
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.
+
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!}{3! \cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88
+
~quacker88
  
 
==Solution 2==
 
==Solution 2==
  
−
We can repeat chooses extensively to find the answer.
 
 
There are <math>7</math> choose <math>3</math> ways to arrange the yellow tiles which is <math>35</math>.
 
There are <math>7</math> choose <math>3</math> ways to arrange the yellow tiles which is <math>35</math>.
 
Then from the remaining tiles there are <math>\binom{4}{2}=6</math> ways to arrange the green tiles.
 
Then from the remaining tiles there are <math>\binom{4}{2}=6</math> 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,
 
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 <math>35*6*2=420</math>
+
giving us an answer of <math>35\cdot6\cdot2=420</math>.
−
<math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math>  
 
−
- noahdavid
 
−
- (edited)starshooter11
 
  
−
==Video Solution==
+
~noahdavid (Edited by starshooter11)
 +
 
 +
==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{\textbf{(B)}\ 420}</math>.
 +
 
 +
~@azure123456 BZ
 +
 
 +
== Solution 4 (Concise)==
 +
 
 +
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>.
 +
 
 +
~MrThinker
 +
 
 +
==Video Solution (HOW TO CRITICALLY THINK!!!)==
 +
https://youtu.be/SQ8XRpl2gXE
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
===Video Solution ==
 +
https://youtu.be/0W3VmFp55cM?t=540
 +
 
 +
===Video Solution 2===
 
https://youtu.be/Gkm5rU5MlOU
 
https://youtu.be/Gkm5rU5MlOU
  
−
~IceMatrix
+
===Video Solution 3===
−
 
 
 
https://youtu.be/4vz8IPfzs2c
 
https://youtu.be/4vz8IPfzs2c
  
 
~savannahsolver
 
~savannahsolver
 +
 +
===Video Solution 4===
 +
https://youtu.be/waxVSOt_v1M?t=275
 +
 +
~ AMBRIGGS
  
 
==See Also==
 
==See Also==

Latest revision as of 13:55, 6 June 2023

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 $10\cdot 6\cdot 7=\boxed{\textbf{(B)}\ 420}$.

~@azure123456 BZ

Solution 4 (Concise)

Let $B$ be brown, $P$ be purple, $G$ be green, and $Y$ be yellow. Then, we are just ordering $Y$, $Y$, $Y$, $G$, $G$, $B$, and $P$. Hence, $\frac{7!}{3! \cdot 2!} = \boxed{\textbf{(B)}\ 420}$.

~MrThinker

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/SQ8XRpl2gXE

~Education, the Study of Everything



=Video Solution

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

Video Solution 2

https://youtu.be/Gkm5rU5MlOU

Video Solution 3

https://youtu.be/4vz8IPfzs2c

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_5&oldid=193992"