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 $\frac{2}{4} = \boxed{\frac{1}{2}}$ .

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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2007 AMC 8 Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

See Also