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Difference between revisions of "2009 AIME I Problems/Problem 9"

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== Problem ==
 
== Problem ==
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A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <dollar/><math>1</math> to <dollar/><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.
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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.
  
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== Solution ==
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== Solution 1 ==
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We are given the seven digits of the values, so let us find the ways we can distribute these digits among the prizes. We are supposed to find the ways to group the digits such that each price has at least one digit, and no prices have five or more digits. We can group the seven digits between the three prices in this way as:
+
[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.]
  
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<math>4</math> digits, <math>2</math> digits, <math>1</math> digit
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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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<math>3</math> digits, <math>3</math> digits, <math>1</math> digit
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<cmath>1131331.</cmath>
  
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<math>3</math> digits, <math>2</math> digits, <math>2</math> digits
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Since the strings have seven digits and three threes, there are <math>\binom{7}{3}=35</math> arrangements of all such strings.
  
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Lets begin with the first case. The possibilities are like so:
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In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.
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When we construct the table for the 1- and 2-digit numbers, we just permute the other four digits to find the choices for the 4-digit number
 
  
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1-digit number      2-digit number                4-digit number
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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
  
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<math>1</math>                    <math>11</math>                        <math>_4C_1 = 4</math>
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<cmath>x+y+z=7, x,y,z>0.</cmath>
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<math>1</math>                    <math>13</math>                        <math>_4C_2 = 6</math>
 
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<math>1</math>                    <math>31</math>                        <math>_4C_2 = 6</math>
 
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<math>1</math>                    <math>33</math>                        <math>_4C_3 = 4</math>
 
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<math>3</math>                    <math>11</math>                        <math>_4C_2 = 6</math>
 
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<math>3</math>                    <math>13</math>                        <math>_4C_3 = 4</math>
 
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<math>3</math>                    <math>31</math>                        <math>_4C_3 = 4</math>
 
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<math>3</math>                    <math>33</math>                        <math>_4C_4 = 4</math>
 
  
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Can the person that did the above solution check out the answer. It is 420
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This gives us
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It seems like you assumed A>B>C, I made the same mistake also on AIME, and where is the other 2 cases
 
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I'll provide mine solution.
 
  
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----
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<cmath>\binom{6}{2}=15</cmath>
  
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'''
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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>.
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Solution2:'''
 
  
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Since we have 3 numbers, consider how many ways we can put this 3 number in a string of 7 digits by putting A,B,C together
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Thus, each arrangement has <cmath>\binom{6}{2}-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways.
  
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For example: <math>A=113, B=13 C=313</math>
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== Solution 1a (Casework)==
  
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Then the string is
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Follow Solution 1 until the partitioning of all the possible strings into 3 groups. Another way to partition the strings is by casework.
  
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<cmath>11313313</cmath>
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Case 1: one of [A, B, or C] has one digit, another has two digits, and the last has four digits.
 +
There are <math>3!=6</math> ways this can happen.
  
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Since the strings have 7 digits and 3 three's. There are
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Case 2: one of [A, B, or C] has two digit, another has two digits, and the last has three digits.
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There are <math>3!/2=3</math> ways this can happen.
  
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<math>_7C_3</math> of such string
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Case 3: one of [A, B, or C] has one digit, another has three digits, and the last has three digits.
 +
There are <math>3!/2=3</math> ways this can happen.
  
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In other to obtain all combination of A,B,C. We partition all the possible strings into 3 groups
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The total numbers of ways per arrangement is <math>6+3+3=12</math> ways. Following Solution 1, there are <math>12\times35=\boxed{420}</math> total ways.
  
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Let look at the example.
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-unhappyfarmer
  
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We have to partition it into 3 groups with each group having at least 1 digit
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==Video Solution==
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https://youtu.be/VhyLeQufKr8 (unavailable)
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We have to find solution where
 
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<cmath>x+y+z=0, 0<x,y,z<5</cmath>
 
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This gives us:
 
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<cmath>_6C_2</cmath> (balls and urns)
 
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But we have counted the one with 5 digit numbers. That is <math>(5,1,1),(1,1,5),(1,5,1)</math>
 
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Thus, each arrangement has <cmath>(_6C_2)-3=12</cmath> ways per arrangement
 
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Thus, there are <math>(12)(35)ways=\boxed{420}</math>
 
  
 
== 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}}

Latest revision as of 12:42, 13 September 2025

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 1

[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$, 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 $A=113, B=13, C=31$, 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 stars and bars. 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.

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 $3!=6$ 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 $3!/2=3$ 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 $3!/2=3$ ways this can happen.

The total numbers of ways per arrangement is $6+3+3=12$ ways. Following Solution 1, there are $12\times35=\boxed{420}$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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