| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Problem from M500 (uk open university maths society) (Dave Rudge).
I was recently checking the pieces of a child's twelve peice jigsaw
puzzle. No two of the first six peices I picked up interlocked -
nor should they have done, since no two of the six selected were
neighbours. The jigsaw puzzle was a standard 4 x 3 rectangular one.
What is the probability of selecting six such pieces?
How about the general case - an m x n = p-piece puzzle, with none
of k selected pieces interlocking, k<=p/2 if p is even, k<=(p+1)/2
if p is odd?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 863.1 | The kids have it easier, they just assemble | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Thu Apr 14 1988 16:20 | 15 |
> I was recently checking the pieces of a child's twelve peice jigsaw > puzzle. No two of the first six peices I picked up interlocked - > nor should they have done, since no two of the six selected were > neighbours. The jigsaw puzzle was a standard 4 x 3 rectangular one. > What is the probability of selecting six such pieces? There are only two ways of selecting the pieces described: if they correspond to either the light or the dark squares of a checkerboard. So the probability is 2/C(12,6) = 2*6!*6!/12! = 1/462. Where you have chosen fewer than half the pieces there are many more possibilities and it gets a lot harder to calculate, because at some points more of the squares of the opposite color become available. Messy. If you have chosen more than half the pieces only one case is possible... | |||||