Imagine you have a 4 x 2 blue lego, a 4 x 2 red lego and a 2 x 2 yellow lego. How many ways, excluding rotations, can you put the legos together (you must use all the pieces)?
Thoughts: With three blocks – one block must be attached to the other two. Not ready to tackle that . . . yet.
Just going to go for two block connections. 74 varieties excluding rotations.
My labeling:
Red Blue Yellow
___ ___ ____
|1 2| |a b| I II
|3 4| |c d| III_IV
|5 6| |e f|
|7_8| |g_h|
Red to Blue:
One connection (4)
1 – a
1 – b
2 – a
2 – b
Two connections (8)
1 2 – a b (respectively)
– a c
– c e
– e g
1 3 – a b
- a c
3 5 – a b
5 7 – a b
Three connections (2)
2 4 6 – c e g
4 6 8 – a c e
Four connections (5)
2 4 6 8 - a c e g
1 2 3 4 – d c b a
- b d a c
- d f c e
- f h e g
Five connections (0)
Six connections (1)
1 2 3 4 5 6 – c d e f g h
Seven connections (0)
Eight connections (1)
1 2 3 4 5 6 7 8 – a b c d e f g h
Now double the total since either the red or the blue block can be on top. 2(4 + 8 + 2 + 5 + 0 + 1 + 0 + 1) = 42
Yellow to Blue (or red)
One connection (2)
1 - I
2 – I
Two connections (4)
1 2 – I II
1 3 – I II
3 5 – I II
5 7 – I II
Three connections (0)
Four connections (2)
1 2 3 4 – I II III IV
3 4 5 6 – I II III IV
Now double the total for top/bottom then double for red or blue connection to yellow 4(2 + 4 + 0 + 2) = 32
Next: Make a 42 by 32 matrix and eliminate the impossible combinations.
Edited on October 5, 2006, 5:03 pm
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Posted by Leming
on 2006-10-05 16:55:18 |