Difference between revisions of "2006 AMC 10A Problems/Problem 18"
m (2006 AMC 10A Problem 18 moved to 2006 AMC 10A Problems/Problem 18) |
(Added solution) |
||
| Line 4: | Line 4: | ||
<math>\mathrm{(A) \ } 10^4\times 26^2\qquad\mathrm{(B) \ } 10^3\times 26^3\qquad\mathrm{(C) \ } 5\times 10^4\times 26^2\qquad\mathrm{(D) \ } 10^2\times 26^4\qquad\mathrm{(E) \ } 5\times 10^3\times 26^3\qquad</math> | <math>\mathrm{(A) \ } 10^4\times 26^2\qquad\mathrm{(B) \ } 10^3\times 26^3\qquad\mathrm{(C) \ } 5\times 10^4\times 26^2\qquad\mathrm{(D) \ } 10^2\times 26^4\qquad\mathrm{(E) \ } 5\times 10^3\times 26^3\qquad</math> | ||
== Solution == | == Solution == | ||
| + | There are <math>10\cdot10\cdot10\cdot10 = 10^4</math> ways to choose 4 digits. | ||
| + | |||
| + | There are <math>26 \cdot 26 = 26^2</math> 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 <math>5</math> choices for position. | ||
| + | |||
| + | Therefore there are <math> 5\times 10^4\times 26^2 </math> distinct license plates <math> \Rightarrow C </math> | ||
| + | |||
== See Also == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 11:21, 17 July 2006
Problem
A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 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 position.
Therefore there are 
distinct license plates 
