Difference between revisions of "1988 AIME Problems/Problem 1"
Diophantient (talk | contribs) (Solution to an AIME problem) |
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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 | ||
− | <math>2^{10}-{10\choose 0}-{10\choose | + | <math>2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=770</math> |
+ | |||
== See also == | == See also == | ||
* [[1988 AIME Problems]] | * [[1988 AIME Problems]] | ||
{{AIME box|year=1988|before=First Question|num-a=2}} | {{AIME box|year=1988|before=First Question|num-a=2}} |
Revision as of 15:26, 27 March 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 {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?
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 |