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

m (Solution)
m (Solution)
Line 9: Line 9:
 
then the string is
 
then the string is
  
<cmath>1131331</cmath>.
+
<cmath>1131331.</cmath>
  
 
Since the strings have seven digits and three threes, there are
 
Since the strings have seven digits and three threes, there are
Line 19: Line 19:
 
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
 
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
  
<cmath>x+y+z=7, 0<x,y,z</cmath>.
+
<cmath>x+y+z=7, x,y,z>0.</cmath>
  
 
This gives us
 
This gives us
Line 25: Line 25:
 
<cmath>_6C_2</cmath>
 
<cmath>_6C_2</cmath>
  
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>.
+
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>.
  
 
Thus, each arrangement has <cmath>(_6C_2)-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways.
 
Thus, each arrangement has <cmath>(_6C_2)-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways.

Revision as of 21:39, 12 March 2012

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

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

$_7C_3$ arrangements of 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

\[_6C_2\]

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 \[(_6C_2)-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways.

See also

2009 AIME I (ProblemsAnswer KeyResources)
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