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

(New page: We are given the 7 digits of the values, so let us find the ways we can distrubute these digits among the prizes. We are supposed to find the ways to group the digits such that each price ...)
 
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We are given the 7 digits of the values, so let us find the ways we can distrubute 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 5 or more digits. We can express 7 this way as :
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== 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 <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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4, 2, 1
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== Solution ==
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3, 3, 1
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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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3, 2, 2
 
  
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Let us begin with the first case. The possibilities are like so:
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<cmath>1131331.</cmath>
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*When we construct the table for the 1 and 2 digit numbers, we just permute the other 4 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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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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1                     11            4 choose 1, or 4 numbers
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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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1                    13            4 choose 2, or 6 numbers
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1                     31            4 choose 2, or 6 numbers
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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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1                     33            4 choose 3, or 4 numbers
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3                    11            4 choose 2, or 6 numbers
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<cmath>x+y+z=7, x,y,z>0.</cmath>
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3                    13            4 choose 3, or 4 numbers
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3                    31            4 choose 3, or 4 numbers
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This gives us
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3                    33            4 choose 4, or 1 number
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--------------------------------------------------------------
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<cmath>\binom{6}{2}=15</cmath>
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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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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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==Video Solution==
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https://youtu.be/VhyLeQufKr8
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== See also ==
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{{AIME box|year=2009|n=I|num-b=8|num-a=10}}
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[[Category:Intermediate Combinatorics Problems]]
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{{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)
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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