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University of South Carolina High School Math Contest/1993 Exam/Problem 8

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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)}.



Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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