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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 8"
 
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What is the coefficient of <math>x^3</math> in the expansion of  
 
What is the coefficient of <math>x^3</math> in the expansion of  
  
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<center><math>4 (1 + x + x^2 + x^3 + x^4 + x^5 )^6? </math></center>
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<center><math> (1 + x + x^2 + x^3 + x^4 + x^5 )^6? </math></center>
  
 
<center><math> \mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64 </math></center>
 
<center><math> \mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64 </math></center>
  
 
== Solution ==
 
== Solution ==
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The expression simplifies to <math>\left(\frac{x^{6}-1}{x-1}\right)^{6}</math>. Expanding both the numerator and denominator, we see that the coefficient of the <math>x^{3}</math> term is <math>{6\choose 5}+{6\choose 3}+{6\choose 6}+{6\choose 3}=56</math>.
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If we expand out the given product, we see that we have a sum of terms in which each term is a product of six members of the set <math>\{1, x, x^2, x^3, x^4, x^5\}</math> (with repetitions allowed).  In order to have one of these terms equal to <math>x^3</math>, we can either have a single <math>x^3</math> term and five terms of 1 in our product (<math>6\choose 1</math> ways) or one <math>x^2</math> term, one <math>x</math> term and four 1 terms (<math>6\cdot 5</math> ways) or have three <math>x</math> terms (<math>6\choose 3</math> ways).  This gives us a total of <math>{6\choose 1} + 6\cdot 5 + {6\choose 3} = 6 + 30 + 20 = 56 \Longrightarrow \mathrm{(C)}</math>.
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Latest revision as of 17:13, 17 August 2006

Problem

What is the coefficient of $x^3$ in the expansion of

$(1 + x + x^2 + x^3 + x^4 + x^5 )^6?$
$\mathrm{(A) \ } 40 \qquad \mathrm{(B) \ }48 \qquad \mathrm{(C) \ }56 \qquad \mathrm{(D) \ }62 \qquad \mathrm{(E) \ } 64$

Solution

If we expand out the given product, we see that we have a sum of terms in which each term is a product of six members of the set $\{1, x, x^2, x^3, x^4, x^5\}$ (with repetitions allowed). In order to have one of these terms equal to $x^3$, we can either have a single $x^3$ term and five terms of 1 in our product ($6\choose 1$ ways) or one $x^2$ term, one $x$ term and four 1 terms ($6\cdot 5$ ways) or have three $x$ terms ($6\choose 3$ ways). This gives us a total of ${6\choose 1} + 6\cdot 5 + {6\choose 3} = 6 + 30 + 20 = 56 \Longrightarrow \mathrm{(C)}$.


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