Difference between revisions of "1988 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
Currently there are <math>{10 \choose 5}</math> possible ways. | Currently there are <math>{10 \choose 5}</math> possible ways. | ||
| − | With any number from 1 to 9, the number of ways is <math>\sum^{9}_{k=1}{10 \choose k}</math>. | + | 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>. |
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 | + | So the number of ways is just |
| − | <math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=770</math> | + | <math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}</math>. |
== See also == | == See also == | ||
Revision as of 20:43, 9 April 2008
Problem
One commercially available ten-button lock may be opened by depressing -- 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 ways.
With any number from 
to 
, the number of ways is 
.
Now we can use the identity 
.
So the number of ways is just
.
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 | ||
