## Wednesday, December 06, 2006

Okay, here is how i would solve the problem below,

When the cube is disassembled, we have 8 cubes with 3 black sides, 12 cubes with 2 black sides, 6 cubes with one black side and 1 cube with none.

Now when the cubes are put back, the 27 cubes can be arranged in 27! ways and each of the cubes can have 24 different orientations. So the denominator would be 24^27 * 27 !

Now going for the numerator.

The cubes with three black sides have to be arranged in the 8 corners and can be arranged in 8! ways and they can each be oriented in 3 ways so that the outerside is black and hence the total number of arrangements = 3^8 * 8!

The cubes with two black sides can be arranged in 12! ways and they can each be oriented in 2 ways so that the outerside is black and hence the total number of arrangements = 2^12 * 12!

The cubes with one black side can be arranged in 6! ways and they can each be oriented in 4 ways so that the outerside is black and hence the total number of arrangements = 4^6 * 6!

Now the last cube , the one with no black face has to go at the center of the cube and can be oriented in 24 ways. So total number of arrangements = 24.

So the probability that the reassembled cube is black
= (3^8 * 8! * 2^12 * 12! * 4^6 * 6! * 24) / (24^27 * 27 !)

Did i miss anything ??