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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 ==
−
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
+
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
  
−
For example: <math>A=113, B=13, C=313</math>
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<cmath>1131331.</cmath>
  
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Then the string is
+
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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<cmath>11313313</cmath>
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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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Since the strings have 7 digits and 3 three's. There are
+
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>_7C_3</math> of such string
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<cmath>x+y+z=7, x,y,z>0.</cmath>
  
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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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This gives us
  
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Let look at the example.
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<cmath>\binom{6}{2}=15</cmath>
  
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We have to partition it into 3 groups with each group having at least 1 digit
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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>.
  
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We have to find solution where
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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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<cmath>x+y+z=7, 0<x,y,z<5</cmath>
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==Video Solution==
−
 
+
https://youtu.be/VhyLeQufKr8
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This gives us:
 
−
 
 
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<cmath>_6C_2</cmath> (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>(_6C_2)-3=12</cmath> ways per arrangement
 
−
 
 
−
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}}

Revision as of 15:56, 20 June 2020

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.

Video Solution

https://youtu.be/VhyLeQufKr8

See also

2009 AIME I (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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