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

m (Solution 2)
m (Solution 2)
Line 16: Line 16:
  
 
We can repeat chooses extensively to find the answer.
 
We can repeat chooses extensively to find the answer.
There are <math>7</math> choose <math>3</math> ways to arrange the brown 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 red 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*6*2=420</math>

Revision as of 19:05, 31 May 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 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.

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

Solution 2

We can repeat chooses extensively to find the answer. 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 $35*6*2=420$ $\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}$ - noahdavid

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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