mwahahaha
2007-06-17 22:27:06 · 3 answers · asked by poisoned_enchantress 1 in Science & Mathematics ➔ Mathematics
10 mines
2007-06-17 22:31:52 · update #1
10 mines
2007-06-17 22:31:53 · update #2
On 81 squares there are 81C10 ways of placing different mines. However the number you want will be less due to mirror images, rotations of the board and improbable minesweeper positions (or minesweeper will not allow all 10 bombs to be in one block with no gaps).
2007-06-17 22:55:09 · answer #1 · answered by blind_chameleon 5 · 0⤊ 0⤋
Depends on how many mines you want to include.
I think, but Im not sure... I never took a probability or combinatorics course... that the solution would be 81Pn, where n is the number of mines.
I say P and not C because the order of the mines dont matter. Combinations are more restrictive than Permutations. But, hey, I warned you... this is where my understanding of combinatorics fails me horrendously.
Anyway... the computed number will be the number as you would perceive them, watching the screen. But if you count layout rotations, mirror images, etc, the true number of possibilities will be greatly reduced.
Furthermore, as someone pointed out, the program/game, itself, may have inherent limitations to how a layout can be "randomly" produced.
2007-06-18 05:30:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If you had 81 squares and n mines to place you would have 81Cn ways to place them.
eg if you had 5 mines, there would be
81C5 = 25621596 ways to place them in the 81 squares.
Just saw your additional detail
with 10 mines,
number of placements = 81C10
=1878392407000 approximately
2007-06-18 05:40:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋