2016 AMC 8 Problems/Problem 17

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

Solution

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. $999 \cdot 10 = 9990 \rightarrow \boxed{D}$

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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2016 AMC 8 Problems/Problem 17

Solution