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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.
  
 
== Solution ==
 
== Solution ==
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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:
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[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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==Video Solution==
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https://youtu.be/VhyLeQufKr8
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Then the string is
 
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<cmath>11313313</cmath>
 
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Since the strings have 7 digits and 3 three's. There are
 
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<math>_7C_3</math> of such string
 
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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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Let look at the example.
 
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We have to partition it into 3 groups with each group having at least 1 digit
 
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We have to find solution where
 
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<cmath>x+y+z=7, 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 22:32, 15 July 2024

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

[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.

Video Solution

https://youtu.be/VhyLeQufKr8

See also

2009 AIME I (Problems • Answer Key • Resources)
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