# 1988 AIME Problems/Problem 1

## Problem

One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?

## Solution

Currently there are ${10 \choose 5}$ possible combinations. With any integer $x$ from $1$ to $9$, the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$. Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$. So the number of additional combinations is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$.

Note: A simpler way of thinking to get $2^{10}$ is thinking that each button has two choices, to be black or

to be white.

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