1982 AHSME Problems/Problem 8

Problem

By definition, $r! = r(r - 1) \cdots 1$ and $\binom{j}{k} = \frac {j!}{k!(j - k)!}$ , where $r,j,k$ are positive integers and $k < j$ . If $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ form an arithmetic progression with $n > 3$ , then $n$ equals

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 12$

Solution

Since $\binom{n}{1}$ , $\binom{n}{2}$ , and $\binom{n}{3}$ form an arithmetic progression, $\binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}$ . Therefore, $\frac{n!}{2!\cdot(n-2)!}-\frac{n!}{1!\cdot(n-1)!}=\frac{n!}{3!\cdot(n-3)!}-\frac{n!}{2!\cdot(n-2)!}.$

Simplifying these expressions yields $\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}$ . Multiplying both sides by $6$ and collecting all terms on one side yields $n^3 - 9n^2 + 14n = 0$ , which factors to $n(n-7)(n-2)=0$ . The solutions to this equation are $n \in \{0, 2, 7\}$ , but $n > 3$ so the only valid answer is $n = \boxed{(\mathbf{B})\ 7}$ .

~ab2024

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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1982 AHSME Problems/Problem 8

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