1988 AIME Problems/Problem 1

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Problem

One commercially available ten-button lock may be opened by depressing -- 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 ways. With any number from $1$ to $9$ , the number of ways 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 ways is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$ .

1988 AIME (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1988 AIME Problems/Problem 1

From AoPSWiki

Problem

Solution

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