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

(→‎Solution)
(→‎Solution)
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I'll provide mine solution.
 
I'll provide mine solution.
  
−
Solution:
+
'''
 +
Solution2:'''
 +
 
 
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 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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Since the strings have 7 digits and 3 three's. There are
 
Since the strings have 7 digits and 3 three's. There are
  
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<math>7 choose 3</math> of such string
+
<math>_7C_3</math> of such string
  
 
In other to obtain all combination of A,B,C. We partition all the possible strings into 3 groups
 
In other to obtain all combination of A,B,C. We partition all the possible strings into 3 groups
Line 54: Line 56:
 
This gives us:
 
This gives us:
  
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<cmath>6 choose 2</cmath> (balls and urns)
+
<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>
 
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>(6 choose 2)-3=12 ways per arrangement</cmath>
+
Thus, each arrangement has <cmath>(_6C_2)-3=12</cmath> ways per arrangement
  
 
Thus, there are <math>(12)(35)ways=\boxed{420}</math>
 
Thus, there are <math>(12)(35)ways=\boxed{420}</math>

Revision as of 20:36, 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

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

Can the person that did the above solution check out the answer. It is 420 It seems like you assumed A>B>C, I made the same mistake also on AIME I'll provide mine solution.

Solution2:

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

For example: $A=113, B=13 C=313$

Then the string is

\[11313313\]

Since the strings have 7 digits and 3 three's. There are

$_7C_3$ of such string

In other to obtain all combination of A,B,C. We partition all the possible strings into 3 groups

Let look at the example.

We have to partition it into 3 groups with each group having at least 1 digit

We have to find solution where

\[x+y+z=0, 0<x,y,z<5\]

This gives us:

\[_6C_2\] (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 \[(_6C_2)-3=12\] ways per arrangement

Thus, there are $(12)(35)ways=\boxed{420}$

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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