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

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== Problem ==
 
== Problem ==
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One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has <math>\{1,2,3,6,9\}</math> 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?
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[[Image:1988-1.png]]
  
 
== Solution ==
 
== Solution ==
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Currently there are <math>{10 \choose 5}</math> possible combinations.
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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>.
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Now we can use the identity <math>\sum^{n}_{k=0}{n \choose k}=2^{n}</math>.
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So the number of additional combinations is just
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<math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}</math>.
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Note: A simpler way of thinking to get <math>2^{10}</math> is thinking that each button has two choices, to be black or
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to be white.
  
 
== See also ==
 
== See also ==
* [[1988 AIME Problems]]
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{{AIME box|year=1988|before=First Question|num-a=2}}
  
{{AIME box|year=1988|before=First Question|num-a=2}}
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[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 18:23, 26 August 2019

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?

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


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.

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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