Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AIME Problems/Problem 1"
(Undo revision 109398 by Nafer (talk))
(Tag: Undo)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
One commercially available ten-button lock may be opened by depressing -- 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?
+
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]]
  
 
== Solution ==
 
== Solution ==
−
Currently there are <math>{10 \choose 5}</math> possible ways.
+
Currently there are <math>{10 \choose 5}</math> possible combinations.
−
With any number from <math>1</math> to <math>9</math>, the number of ways 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>.
 +
 +
 +
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 ==

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_1&oldid=109399"