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

We know that $n \choose {2}$ $-$ $n \choose 1$ $=$ ${n} \choose 3$ $-$ ${n} \choose 2,$ because they form an arithmetic sequence, and expanding, we have by the definitions in the problem: $\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}-n=\frac{n(n-1)(n-2)(n-3)(n-4)...}{6((n-3)(n-4)...)}-\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}.$

Canceling out the $(n-2)!$ and the $(n-3)!$ from each side of the equals sign, we have $\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}.$ Getting rid of the fractions by cross multiplication, and getting n on one side, we have $n^3 - 9n^2 + 14n = 0,$ and we can factor out the n, so n(n^2-9n+14)=0, and we are looking for two integers x and y such that $x+y=-9$ and $xy=14.$ By guess and check, our integers are -7 and -2, so $n(n-7)(n-2)=0!!!$ According to the problem, $n>3,$ so we have n=7 or 2, thus $\boxed{\left(B\right)n=7}$

~ab2024

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1982 AHSME Problems/Problem 8

Problem

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