Track / Stage 3 / 118 of 260 #118 of 2000

Problem 118

AMC 10/12, early questions

Combinatorics Difficulty 3.4 Prove it The 4th Japanese Junior Mathematical Olympiad · Japan

There are ten red cards, numbered $1$ , $2$ , $\ldots$ , $10$ , and ten blue cards, also numbered $1$ , $2$ , $\ldots$ , $10$ . How many ways are there to choose three from these twenty cards so that the sum of the numbers on the cards chosen is $16$ or less?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Write $1$ , $2$ , $\ldots$ , $9$ or $10$ on the back of the cards so that each card has two numbers which add up to $11$ .

Call a set of three cards *good* if the sum of the numbers on their faces is $16$ or less, and call it *bad* if the sum of the numbers on their backs is $16$ or less.

Since the sum of their faces and their backs add up to $11 \times 3 = 33$ , every set of three cards is good or bad, and none is both. Furthermore, by symmetry, there must be the same number of good and bad sets. Therefore, there are exactly $\frac{1}{2} \cdot \binom{20}{3} = 570$ ways of required choice.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.