# 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 $1$ to $9999$ 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 $1, 1, 1, 1, 3, 3, 3$. 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 $A=113, B=13, C=31$, then the string is $$1131331.$$

Since the strings have seven digits and three threes, there are $\binom{7}{3}=35$ 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 $$x+y+z=7, x,y,z>0.$$

This gives us $$\binom{6}{2}=15$$

ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$.

Thus, each arrangement has $$\binom{6}{2}-3=12$$ ways per arrangement, and there are $12\times35=\boxed{420}$ ways.

## See also

 2009 AIME I (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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