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Difference between revisions of "2009 AIME I Problems/Problem 9"
(→‎Solution)
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== Solution ==
 
== Solution ==
−
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 :
+
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:
  
−
4, 2, 1
+
<math>4</math> digits, <math>2</math> digits, <math>1</math> digit
−
3, 3, 1
 
−
3, 2, 2
 
  
−
Let us begin with the first case. The possibilities are like so:
+
<math>3</math> digits, <math>3</math> digits, <math>1</math> digit
−
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
 
  
−
1 digit number        2 digit number          4 digit number
+
<math>3</math> digits, <math>2</math> digits, <math>2</math> digits
  
−
1                    11             4 choose 1, or 4 numbers
+
Lets begin with the first case. The possibilities are like so:
−
1                    13             4 choose 2, or 6 numbers
+
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
−
1                    31             4 choose 2, or 6 numbers
+
 
−
1                    33             4 choose 3, or 4 numbers
+
1-digit number      2-digit number                4-digit number
−
3                    11             4 choose 2, or 6 numbers
+
 
−
3                    13             4 choose 3, or 4 numbers
+
<math>1</math>                     <math>11</math>                        <math>_4C_1 = 4</math>
−
3                    31             4 choose 3, or 4 numbers
+
<math>1</math>                     <math>13</math>                        <math>_4C_2 = 6</math>
−
3                    33             4 choose 4, or 1 number
+
<math>1</math>                     <math>31</math>                        <math>_4C_2 = 6</math>
 +
<math>1</math>                     <math>33</math>                        <math>_4C_3 = 4</math>
 +
<math>3</math>                     <math>11</math>                        <math>_4C_2 = 6</math>
 +
<math>3</math>                     <math>13</math>                        <math>_4C_3 = 4</math>
 +
<math>3</math>                     <math>31</math>                        <math>_4C_3 = 4</math>
 +
<math>3</math>                     <math>33</math>                        <math>_4C_4 = 4</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}}

Revision as of 20:26, 20 March 2009

Problem

A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <dollar/>$1$ to <dollar/>$9999$ 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

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:

$4$ digits, $2$ digits, $1$ digit

$3$ digits, $3$ digits, $1$ digit

$3$ digits, $2$ digits, $2$ digits

Lets begin with the first case. The possibilities are like so: 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

1-digit number 2-digit number 4-digit number

$1$ $11$ $_4C_1 = 4$ $1$ $13$ $_4C_2 = 6$ $1$ $31$ $_4C_2 = 6$ $1$ $33$ $_4C_3 = 4$ $3$ $11$ $_4C_2 = 6$ $3$ $13$ $_4C_3 = 4$ $3$ $31$ $_4C_3 = 4$ $3$ $33$ $_4C_4 = 4$

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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