Maths Olympiad Prep

Track / Stage 4 / 76 of 340 #336 of 1964

Problem 336

AMC 12 late, AIME early
Combinatorics Difficulty 4.8 Multiple choice

11911 \cdot 9 As shown, Pascal's (Yang Hui's) Triangle is an array of positive integers, where the first row is 1, the second row is two 1s, each row starts and ends with 1, and in any row, the kk-th number (when it is not 1) is the sum of the kk-th and (k1)(k-1)-th numbers in the previous row. In the first nn rows, the ratio of the number of numbers that are not 1 to the number of 1s is
111121133114641 \begin{array}{ccccc} & & 1 & & \\ & 1 & 1 & & \\ & 1 & 2 & 1 & \\ & 1 & 3 & 3 & 1 \\ 1 & 4 & 6 & 4 & 1 \end{array}

Pick one

Official solution

[Solution] In the first nn rows, the number of 1's is (2n1)(2 n-1), and the number of other numbers is
12(n2)(n1)=12(n23n+2) .  \frac{1}{2}(n-2)(n-1)=\frac{1}{2}\left(n^{2}-3 n+2\right) \text { . }

The quotient is
n23n+24n2 \frac{n^{2}-3 n+2}{4 n-2} \text {. }

Therefore, the answer is (D)(D).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.