1988 AIME Problems/Problem 1: Difference between revisions
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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 1 to 9, 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 the difference in the number of ways is just | So the difference in the number of ways is just | ||
Revision as of 11:50, 24 October 2007
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?
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Solution
Currently there are 
possible ways.
With any number from 1 to 9, the number of ways is 
.
Now we can use the identity 
.
So the difference in 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 | ||
