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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 20"
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We have <math>(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1</math> (The <math>-1</math> since we have one less set). This is <math>7!-1=5039</math>.
 
We have <math>(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1</math> (The <math>-1</math> since we have one less set). This is <math>7!-1=5039</math>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 19|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 21|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 10:53, 23 July 2006

Problem

Let $A_1, A_2, \ldots , A_{63}$ be the 63 nonempty subsets of $\{ 1,2,3,4,5,6 \}$. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_i$. Then what is the value of $\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})$?

$\mathrm{(A) \ }5003 \qquad \mathrm{(B) \ }5012 \qquad \mathrm{(C) \ }5039 \qquad \mathrm{(D) \ }5057 \qquad \mathrm{(E) \ }5093$

Solution

We have $(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)-1$ (The $-1$ since we have one less set). This is $7!-1=5039$.

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