By definition r!=r(r-1)…1 and jk=(j!)/(k!(j-k)!), where r, j, k are positive integers and k 3, then n equals

1982 AMC 12 Problem 8

Attempt Problem 8 of the 1982 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1982 AMC 12 solutions, or check the answer key.

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By definition

r!=r(r-1)\cdots1

and

\binom jk=\frac{j!}{k!(j-k)!},

where

r,

j,

k

are positive integers and

k\lt j.

\binom n1,

\binom n2,

\binom n3

form an arithmetic progression with

n\gt3,

then

n

equals

5

7

9

11

12

The middle term of an arithmetic progression is the average of its neighbors

Substitute the factorial formulas into

2\binom n2=\binom n1+\binom n3

The progression condition is

2\binom n2=\binom n1+\binom n3.

Substitution and cancellation give

n^2-9n+14=0,

(n-2)(n-7)=0.

Since

n\gt3,

n=7.

Therefore, the correct answer is B.