Difference between revisions of "2020 AMC 10B Problems/Problem 5"
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<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> | ||
− | ==Solution== | + | ==Solution 1== |
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>. |
− | + | ~quacker88 | |
− | ==Solution 2 | + | ==Solution 2== |
− | + | 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. | |
+ | 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\cdot6\cdot2=420</math>. | ||
− | ==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{420}.</math> | ||
+ | |||
+ | ~@azure123456 BZ | ||
+ | |||
+ | == Video Solutions == | ||
+ | ===Video Solution 1=== | ||
+ | https://youtu.be/0W3VmFp55cM?t=540 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | ===Video Solution 2=== | ||
https://youtu.be/Gkm5rU5MlOU | https://youtu.be/Gkm5rU5MlOU | ||
~IceMatrix | ~IceMatrix | ||
+ | ===Video Solution 3=== | ||
+ | https://youtu.be/4vz8IPfzs2c | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 18:29, 6 September 2021
Contents
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.)
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 ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities: .
~quacker88
Solution 2
There are choose ways to arrange the yellow tiles which is . Then from the remaining tiles there are 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 .
~noahdavid (Edited by starshooter11)
Solution 3
We can choose a different frame to solve this problem. Our tile combination can be written as We can focus on first, which gives us Now we can insert our brown tile into this which only has choices(like and etc.), then insert purple tile which only has choices(like and etc.). Multiply them together we get
~@azure123456 BZ
Video Solutions
Video Solution 1
https://youtu.be/0W3VmFp55cM?t=540
~ pi_is_3.14
Video Solution 2
~IceMatrix
Video Solution 3
~savannahsolver
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.