Difference between revisions of "2009 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | [Clarification: You are supposed to find the number of all possible tuples of prices, <math>(A, B, C)</math>, 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 <math>A=113, B=13, C=31</math>, then the string is | 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 <math>A=113, B=13, C=31</math>, then the string is | ||
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<cmath>\binom{6}{2}=15</cmath> | <cmath>\binom{6}{2}=15</cmath> | ||
− | ways by | + | ways by stars and bars. 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}-3=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. |
Latest revision as of 23:32, 15 July 2024
Contents
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
[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
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
This gives us
ways by stars and bars. But we have counted the one with 5 digit numbers; that is, .
Thus, each arrangement has ways per arrangement, and there are
ways.
Video Solution
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.