==> 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 /
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