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On the probability of placing cups on saucers


 * A tea set consists of six cups and saucers with two cups and saucers is each of the three different colours. The cups are placed randomly on the saucers. What is the probability that no cups is on a saucer of the same colour?

What we do need here is a painting factory.

The two associated species equation are :

$$F = 1 + (R+G+B)\cdot (r+g+b) \cdot F$$  and

$$F = 1 + (Rr+Rg+Br+Bg+Gr+Gb)\cdot F $$

The generating function for the first "production line" is: $$f(x_R,x_G,x_B,x_r,x_g,x_b) = {1 \over 1-(x_R+x_G+x_B)(x_r+x_g+x_b)} $$

By Maple, the coefficient of $$x_R^2x_G^2x_B^2x_r^2x_g^2x_b^2$$ is $$8100$$ and it represents the number of all possible paintings of six china sets under the initial restrictions of color occurences.

The generating function for the second "production line" is: $$f(x_R,x_G,x_B,x_r,x_g,x_b) = {1 \over 1 - (x_Rx_r+x_Rx_g+x_Bx_r+x_Bx_g+x_Gx_r+x_Gx_b)} $$

By taking into account the second coefficient we obtain:

$$P= {900 \over 8100} = {1 \over 9}$$