Difference between revisions of "2020 AMC 10B Problems/Problem 5"
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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== | ||
+ | |||
+ | 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>. | ||
+ | Then from the remaining tiles there are <math>4</math> choose <math>2=6</math> ways to arrange the red 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> | ||
+ | <math>\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}</math> | ||
+ | - noahdavid | ||
==Video Solution== | ==Video Solution== |
Revision as of 18:03, 11 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.)
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 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
We can repeat chooses extensively to find the answer. There are choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red 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
Video Solution
~IceMatrix
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.