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 | ||
− | <center><math> | + | <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 == | ||
− | + | 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 in the expansion of
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 (with repetitions allowed). In order to have one of these terms equal to , we can either have a single term and five terms of 1 in our product ( ways) or one term, one term and four 1 terms ( ways) or have three terms ( ways). This gives us a total of .