# Difference between revisions of "2020 AMC 10B Problems/Problem 17"

The following problem is from both the 2020 AMC 10B #17 and 2020 AMC 12B #15, so both problems redirect to this page.

## Problem

There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other? $\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$

## Solution

Let us use casework on the number of diagonals.

Case 1: $0$ diagonals There are $2$ ways: either $1$ pairs with $2$, $3$ pairs with $4$, and so on or $10$ pairs with $1$, $2$ pairs with $3$, etc.

Case 2: $1$ diagonal There are $5$ possible diagonals to draw (everyone else pairs with the person next to them.

Note that there cannot be 2 diagonals.

Case 3: $3$ diagonals There are $5% possible diagonals to draw. Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction. Case 4:$ (Error compiling LaTeX. ! Missing $inserted.)5 $diagonals There is$1$way to do this.

Thus, in total there are$(Error compiling LaTeX. ! Missing$ inserted.)2+5+5+1=\boxed{13}\$ possible ways.

## Video Solution

https://youtu.be/3BvJeZU3T-M (for AMC 10) https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 