Difference between revisions of "1988 AIME Problems/Problem 1"
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So the number of additional combinations 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>. | ||
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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 | ||
| + | |||
| + | to be white. | ||
== See also == | == See also == | ||
Revision as of 15:10, 28 June 2017
Problem
One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has 
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 
possible combinations.
With any integer 
from 
to 
, the number of ways to choose a set of buttons is 
.
Now we can use the identity 
.
So the number of additional combinations is just
.
Note: A simpler way of thinking to get 
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. 
