Difference between revisions of "1988 AIME Problems/Problem 1"

m (wik)
m (Solution)
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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 12:50, 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 $\{1,2,3,6,9\}$ 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?

1988-1.png

Solution

Currently there are ${10 \choose 5}$ possible ways. With any number from 1 to 9, the number of ways is $\sum^{9}_{k=1}{10 \choose k}$. Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$. So the difference in the number of ways is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=770$

See also

1988 AIME (ProblemsAnswer KeyResources)
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
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