Difference between revisions of "2020 AMC 10B Problems/Problem 5"
(→Solution) |
m (→Solution) |
||
| Line 9: | Line 9: | ||
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. | + | 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 18:00, 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.)
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
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. 
