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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 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?
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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?
 +
 
 
[[Image:1988-1.png]]
 
[[Image:1988-1.png]]
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== Solution ==
 
== Solution ==
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Currently there are <math>{10 \choose 5}</math> possible ways.
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Currently there are <math>{10 \choose 5}</math> possible combinations.
−
With any number from 1 to 9 the number of ways is <math>\sum^{9}_{k=1}{10 \choose k}</math>.
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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>.
 
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 difference in 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=770</math>
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<math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}</math>.
 +
 
 +
 
 +
Note: A simpler way of thinking to get <math>2^{10}</math> is thinking that each button has two choices, to be black or
 +
 +
to be white.
  
 
== See also ==
 
== See also ==
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* [[1988 AIME Problems]]
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{{AIME box|year=1988|before=First Question|num-a=2}}
  
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{{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 17:23, 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 (Problems • Answer Key • Resources)
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

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

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