# 2007 AMC 8 Problems/Problem 24

## Problem

A bag contains four pieces of paper, each labeled with one of the digits $1$, $2$, $3$ or $4$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$? $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$

## Solution 1

The number of ways to form a 3-digit number is $4 \cdot 3 \cdot 2 = 24$. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is $3! + 3! = 12$. Therefore, the probability is $\frac{12}{24} = \boxed{\frac{1}{2}}$.

~abc2142

## Solution 2

Only (1,2,3) and (2,3,4) are possible combinations that sum to 3. The total ways to choose three numbers from the four numbers is 4 choose 3 which equals 4. The answer is $\frac{2}{4} = \boxed{\frac{1}{2}}$.

## Video Solution

https://youtu.be/hwc11K02cEc - Happytwin

## Video Solution 2

https://youtu.be/cUxFS9l-Pb4 - Soo, DRMS, NM

 2007 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions