2016 AMC 8 Problems/Problem 17

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An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$

Solution 1

For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}$

Solution 2

Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1,1 and we can have any of the 10 digits for the last digit. So our answer is $10^4-10=\boxed{\textbf{(D)}\ 9990}.$

 2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 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