2012 AMC 8 Problems/Problem 10

Problem

How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$

Solution 1

For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$ , since all of the valid 4-digit number will always be greater than $1000$ . The best way to solve this problem is by using casework.

There can be only two leading digits, namely $1$ or $2$ .

When the leading digit is $1$ , you can make $\frac{3!}{2!1!} \implies 3$ such numbers.

When the leading digit is $2$ , you can make $3! \implies 6$ such numbers.

Summing the amounts of numbers, we find that there are $\boxed{\textbf{(D)}\ 9}$ such numbers.

Solution 2

Notice that the first digit cannot be $0$ , as the number is greater than $1000$ . Therefore, there are three digits that can be in the thousands.

The rest three digits of the number have no restrictions, and therefore there are $3! \implies 6$ for each leading digit.

Since the two $2$ 's are indistinguishable, there are $\frac {3\cdot6}{2}$ such numbers $\implies \boxed{\textbf{(D)}\ 9}$ .

Solution 3(a simpler version of Solution 2)

Video Solution

https://youtu.be/OBKIeTgw4Zg ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

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

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