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

Revision as of 14:37, 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

Solution 2 (2 second solve)

Be a memer and guess $\boxed{\textbf{(B) }420}$ , which happens to be the correct answer. -fidgetboss_4000

Video Solution

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 ==Solution 2 (2 second solve)==
-Assume that MAA is an immature bunch and guess <math>\boxed{\textbf{(B) }420}</math>, which happens to be the correct answer.
+Be a memer and guess <math>\boxed{\textbf{(B) }420}</math>, which happens to be the correct answer.
 -fidgetboss_4000

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