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

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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 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!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88
 
<math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> ~quacker88

Revision as of 17:00, 7 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

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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