how many combinations in a list equal a certain total
Quibbles.
With 479 items, there are 2^479-1 combinations to check.
Yes, don't know what I was thinking - thanks for the correction
8,640,000 secons is 100 days, not years.
Already admitted to that one
Have to think about your maximum combinations assertion - (not questioning
it) - thanks
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Regards,
Tom Ogilvy
"Harlan Grove" wrote in message
...
"Tom Ogilvy" wrote...
There are 2^480 - 1 unique combinations to check. As a point of
comparison, there are only 8,640,000 seconds in 100 years (slightly
larger
then 2^23) . . .
Quibbles.
With 479 items, there are 2^479-1 combinations to check.
8,640,000 secons is 100 days, not years.
. . . In otherwords, there isn't enough time to do an exhaustive
examination without even trying to look at the results.
You could come up with a list of Unique prices and reduce the number of
rows
you need to examine. (so in your example, any solution that contains a B
could be another solution by replacing B with F)
Add to those heuristics throwing out all individual items with prices over
$100.
Also, for k = INT(100/MIN(PriceList)), k gives the largest cardinality
combinations that need to be checked. That means only 479 choose k
combinations
to check. Still, this type of problem lacks practical means to achieve a
comprehensive solution.
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