# 2008 Mock ARML 2 Problems/Problem 3

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## Problem

A variation of Pascal's triangle is constructed by writing the numbers $2$ and $3$ in the top row and writing each subsequent term as the sum of the two terms above it. Find the fifth term from the left in the thirteenth row. $$\begin{tabular}{ccccccccccc} & & & & & & & & & & \\ & & & & 2 & & 3 & & & & \\ & & & 2 & & 5 & & 3 & & & \\ & & 2 & & 7 & & 8 & & 3 & & \\ & 2 & & 9 & & 15 & & 11 & & 3 & \\ 2 & & 11 & & 24 & & 26 & & 14 & & 3 \\ \end{tabular}$$

## Solution

Note that this variation can be constructed using overlapping Pascal's triangles. Consider multiplying each of the terms in Pascal's triangles by $2$; then taking another triangle and multiplying each of the terms by $3$, shifting the resulting triangle one space to the right, and then summing the overlapping entries. Note that this preserves the desired recursion.

It follows that the fifth term in the thirteenth row is $2{12 \choose 4} + 3{12 \choose 3} = \boxed{1650}$.