Difference between revisions of "1988 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | One commercially available ten-button lock may be opened by | + | 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? |
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[[Image:1988-1.png]] | [[Image:1988-1.png]] | ||
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== Solution == | == Solution == | ||
− | Currently there are <math>{10 \choose 5}</math> possible | + | Currently there are <math>{10 \choose 5}</math> possible combinations. |
− | With any | + | 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 | + | So the number of additional combinations 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=\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 | ||
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+ | to be white. | ||
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== See also == | == See also == | ||
− | + | {{AIME box|year=1988|before=First Question|num-a=2}} | |
− | {{ | + | [[Category:Intermediate Combinatorics Problems]] |
+ | {{MAA Notice}} |
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 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.