Combinatorics AIME Intermediate
1992


Problem - 2707
In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

It is equivalent to solving $$\begin{array}{rl} & C_n^k : C_n^{k+1} : C_n^{k+2} = 3: 4:5 \\ \\ \implies & \frac{n!}{k!(n-k)!} : \frac{n!}{(k+1)!(n-k-1)!} : \frac{n!}{(k+2)!(n-k-2)!}=3:4:5 \\ \implies & \left\{\begin{array}{rcl} (n-k) : (k+1) &=& 4:3 \\ (n-k-1):(k+2) &=& 5:4 \end{array}\right. \\ \\ \implies & n = \boxed{62}, k =26\end{array}$$

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