Difference between revisions of "1988 AIME Problems/Problem 1"

(Solution)
(Solution)
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With any integer <math>x</math> from <math>1</math> to <math>9</math>, the number of ways to choose a set of <math>x</math> buttons is <math>\sum^{9}_{k=1}{10 \choose k}</math>.
 
With any integer <math>x</math> from <math>1</math> to <math>9</math>, the number of ways to choose a set of <math>x</math> buttons is <math>\sum^{9}_{k=1}{10 \choose k}</math>.
 
Now we can use the identity <math>\sum^{n}_{k=0}{n \choose k}=2^{n}</math>.
 
Now we can use the identity <math>\sum^{n}_{k=0}{n \choose k}=2^{n}</math>.
So the number of ways is just
+
So the number of additional combinations is just
 
<math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}</math>.
 
<math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}</math>.
  

Revision as of 01:18, 6 June 2014

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?

1988-1.png

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 $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$.

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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