2009 AIME I Problems/Problem 9
Problem
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from 
to 
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 1
[Clarification: You are supposed to find the number of all possible tuples of prices, 
, that could have been on that day.]
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 stars and bars. 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.
Solution 1a (Casework)
Follow Solution 1 until the partitioning of all the possible strings into 3 groups. Another way to partition the strings is by casework.
Case 1: one of [A, B, or C] has one digit, another has two digits, and the last has four digits.
There are 
ways this can happen.
Case 2: one of [A, B, or C] has two digit, another has two digits, and the last has three digits.
There are 
ways this can happen.
Case 3: one of [A, B, or C] has one digit, another has three digits, and the last has three digits.
There are ways this can happen.
The total numbers of ways per arrangement is 
ways. Following Solution 1, there are total ways.
-unhappyfarmer
Video Solution
https://youtu.be/VhyLeQufKr8 (unavailable)
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
