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Difference between revisions of "2003 AIME I Problems/Problem 3"
m (→‎Solution)
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== Solution ==
 
== Solution ==
−
Each element will appear in <math>7</math> two element subsets. (Once with each other number.)
+
Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.
  
 
<math>34</math> will be the greater number in <math>7</math> subsets.  
 
<math>34</math> will be the greater number in <math>7</math> subsets.  
Line 21: Line 21:
 
<math>1</math> will be the greater number in <math>0</math> subsets.  
 
<math>1</math> will be the greater number in <math>0</math> subsets.  
  
−
Therefore the sum is:
+
Therefore the desired sum is:
  
 
<math> \displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484</math>
 
<math> \displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484</math>

Revision as of 17:25, 4 August 2006

Problem

Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.

Solution

Each element of the set will appear in $7$ two-element subsets, once with each other number.

$34$ will be the greater number in $7$ subsets.

$21$ will be the greater number in $6$ subsets.

$13$ will be the greater number in $5$ subsets.

$8$ will be the greater number in $4$ subsets.

$5$ will be the greater number in $3$ subsets.

$3$ will be the greater number in $2$ subsets.

$2$ will be the greater number in $1$ subsets.

$1$ will be the greater number in $0$ subsets.

Therefore the desired sum is:

$\displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484$

See also

Template:Comb

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