Difference between revisions of "2009 AIME I Problems/Problem 9"
(→Solution) |
|||
| (5 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from | + | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <math>\$ 1</math> to <math>\$ 9999</math> 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 <math>1, 1, 1, 1, 3, 3, 3</math>. Find the total number of possible guesses for all three prizes consistent with the hint. |
== 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 | ||
<cmath>1131331.</cmath> | <cmath>1131331.</cmath> | ||
| − | Since the strings have seven digits and three threes, there are <math> | + | Since the strings have seven digits and three threes, there are <math>\binom{7}{3}=35</math> arrangements of all such strings. |
In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups. | In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups. | ||
| Line 17: | Line 19: | ||
This gives us | This gives us | ||
| − | <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> | + | Thus, each arrangement has <cmath>\binom{6}{2}-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways. |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/VhyLeQufKr8 | ||
== 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}} | ||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 02:11, 31 August 2022
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
![\[1131331.\]](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.
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. 
