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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

389 pickover/pickover.15.p


Title: Cliff Puzzle 15: Cherries in Wine Glasses
From: cliff@watson.ibm.com

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover

* * *

Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries
at the grid. A glass is considered occupied if it contains at least one
cherry. (With each throw a cherry goes into one of the glasses.) How
many different patterns of occupied glasses can you make? (A glass with
more than one cherry is considered the same as a glass with one cherry
in the pattern).

2. Same as above except that you place 8 cherries in glasses (x,y) and
then determine the other positions by placing cherries at (x,-y),
(-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array
of glasses centered at the origin. How many different patterns of
occupied glasses can you make? (A glass with more than one cherry is
considered the same as a glass with one cherry in the pattern).

3. Can your results be extrapolated to an NxN grid with M cherries
thrown at it for both problems?

pickover/pickover.15.s

In article <1992Oct30.173903.108937@watson.ibm.com> you write:
: Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries
: at the grid. A glass is considered occupied if it contains at least one
: cherry. (With each throw a cherry goes into one of the glasses.) How
: many different patterns of occupied glasses can you make? (A glass with
: more than one cherry is considered the same as a glass with one cherry
: in the pattern).
Assuming that rotated patterns are allowed, then it is (simply)
sum( 81!/(81-n)! , n=1->32) . Since, if a total of n different classes are
filled, then the number of combinations is 81!/(81-n)!. Since there can
be from 1 to 32 glasses filled, the total # is just the sum of these...

:
: 2. Same as above except that you place 8 cherries in glasses (x,y) and
: then determine the other positions by placing cherries at (x,-y),
: (-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array
: of glasses centered at the origin. How many different patterns of
: occupied glasses can you make? (A glass with more than one cherry is
: considered the same as a glass with one cherry in the pattern).
This limitation basically reduces the number of available spots, from 9x9
to 5x5. Also, I only have to worry about 8 occupied spaces. Soo...
#of comb. = sum( (25!/(25-n)!, n=1->8)
:
: 3. Can your results be extrapolated to an NxN grid with M cherries
: thrown at it for both problems?
With a odd N, and M = 4k (evenly divs by 4), then
for 1....
#of comb = sum( (N^2)!/(N^2-n)! , n=1->M)
for 2....
#of comb = sum( (((N+1)/2)^2)!/(((N+1)/2)^2-n)! , n=1->M/4)

--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         mneylon@jupiter.cse.utoledo.edu           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L

 

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