Difference between revisions of "2009 AIME I Problems/Problem 9"
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This gives us | This gives us | ||
| − | <cmath>\binom{6}{2}</cmath> | + | <cmath>\binom{6}{2}=15</cmath> |
ways by balls and urns. But we have counted the one with 5 digit numbers; that is, <math>(5,1,1),(1,1,5),(1,5,1)</math>. | ways by balls and urns. But we have counted the one with 5 digit numbers; that is, <math>(5,1,1),(1,1,5),(1,5,1)</math>. | ||
| − | Thus, each arrangement has <cmath>\binom{6}{2}=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways. | + | Thus, each arrangement has <cmath>\binom{6}{2}-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways. |
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=8|num-a=10}} | {{AIME box|year=2009|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:25, 4 March 2014
Problem
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <dollar/>
to <dollar/>
inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were 
. Find the total number of possible guesses for all three prizes consistent with the hint.
Solution
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if 
, then the string is
![\[1131331.\]](http://latex.artofproblemsolving.com/e/8/a/e8a6fbd35349a2cea1203e4a7320d88a7e80378b.png)
Since the strings have seven digits and three threes, there are 
arrangements of all such strings.
In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.
Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to
![\[x+y+z=7, x,y,z>0.\]](http://latex.artofproblemsolving.com/9/5/e/95e226e82aac8471ee273f7599f82d43e97c6bc6.png)
This gives us
![\[\binom{6}{2}=15\]](http://latex.artofproblemsolving.com/7/0/4/704434a4f2ca296aae694cc675568f7111851674.png)
ways by balls and urns. But we have counted the one with 5 digit numbers; that is, 
.
Thus, each arrangement has ![\[\binom{6}{2}-3=12\]](http://latex.artofproblemsolving.com/0/f/9/0f96290a236307f577422607af892165e6c1e467.png)
ways per arrangement, and there are 
ways.
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
