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

(Undo revision 117586 by Fidgetboss 4000 (talk))
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(Undo revision 117585 by Fidgetboss 4000 (talk))
(Tag: Undo)
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<math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88
 
<math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88
 
==Solution 2 (2 second solve)==
 
 
Be a memer and guess <math>\boxed{\textbf{(B) }420}</math>, which happens to be the correct answer.
 
-fidgetboss_4000
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:38, 10 February 2020

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

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.

$\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}$ ~quacker88

Video Solution

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

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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