Difference between revisions of "2006 AMC 10A Problems/Problem 18"
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So, there are <math>6 - 1 = 5</math> choices for the position of the letters. | So, there are <math>6 - 1 = 5</math> choices for the position of the letters. | ||
| − | Therefore, the number of distinct license plates is <math> 5\times 10^4\times 26^2 \Longrightarrow \mathrm{C}</math>. | + | Therefore, the number of distinct license plates is <math> 5\times 10^4\times 26^2 \Longrightarrow \boxed{\mathrm{C}}</math>. |
| − | == Video Solution == | + | |
| + | == Solution 2 (Answer choices) == | ||
| + | There are <math>10^4</math> ways to choose the digits and <math>26^2</math> ways to choose the 2 letters. However, since the letters are next to each other, the result must be <math>10^4\cdot 26^2 \cdot \text{something}</math>. The only answer choice that matches this is <math>\boxed{\mathrm{C}}</math>. | ||
| + | |||
| + | ~ Andrew2019 | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/0W3VmFp55cM?t=847 | ||
| + | |||
| + | == Video Solutions == | ||
https://youtu.be/3MiGotKnC_U?t=1446 | https://youtu.be/3MiGotKnC_U?t=1446 | ||
~ThePuzzlr | ~ThePuzzlr | ||
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https://youtu.be/f56aG55oG8w | https://youtu.be/f56aG55oG8w | ||
Latest revision as of 14:33, 30 July 2024
Contents
Problem
A license plate in a certain state consists of 
digits, not necessarily distinct, and 
letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?
Solution
There are 
ways to choose 4 digits.
There are 
ways to choose the 2 letters.
For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters.
So, there are 
choices for the position of the letters.
Therefore, the number of distinct license plates is 
.
Solution 2 (Answer choices)
There are 
ways to choose the digits and 
ways to choose the 2 letters. However, since the letters are next to each other, the result must be 
. The only answer choice that matches this is 
.
~ Andrew2019
Video Solution by OmegaLearn
https://youtu.be/0W3VmFp55cM?t=847
Video Solutions
https://youtu.be/3MiGotKnC_U?t=1446
~ThePuzzlr
~savannahsolver
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
