picking two same colored balls at the time 
There are 5 white and 5 black balls in a bin. We take two balls at the time without replacement. Describe X = number of takes where both balls have the same color.

I realized that the minimal number of takes will be 0, and the maximum 5. I started describing for $X=0$:
$$P=5/10 * 5/9 * 4/8 * 4/7 * 3/6 * 3/5 * 2/4 * 2/3 * 1/2 * 1/1$$
However, the answer should be 8/63. What did I do wrong? I can't proceed to other cases of X because I do not know the easiest one. If you could point where I made mistake, as well as show me one of other cases of X I would be grateful. Also, is it right in general case that if we pick n balls, that that is the same as we were picking one ball at the time, n times, divided by $n!$ because order does not matter? 

In every turn, you pick two balls. The color of the first ball does not matter, as long as the second ball has a different color. As such, we get:
$$P[X=0] = \frac{5}{9} \cdot \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \cdot \frac{1}{1} = \frac{120}{945} = \frac{8}{63}$$
It is not possible to have $X = 1$, since if this would be true, we would be left with 3 balls of this color and 5 balls of the other. As such, at least one of the next four picks will have two balls of the same color as well. The same is true for $X=3$, and $X=5$ cannot be true since there are an odd number of balls with the same color. As such, we only need to check for $X=2$ and $X=4$. In the former case, we can choose three picks out of five which contain the balls of different color. The probability of the first three picks being picks of different color equals $\frac{5}{9} \cdot \frac{4}{7} \cdot \frac{3}{5}$, so we get:
$$P[X=2] = {5 \choose 3} \cdot \frac{5}{9} \cdot \frac{4 }{7} \cdot \frac{3}{5} \cdot \frac{1}{3} \cdot \frac{1}{1} = 10 \cdot \frac{4}{63} = \frac{40}{63}$$
In the latter case, we can choose one pick which contains two balls of different color. Then, the remaining picks must each contain two balls of the same color. As such, we get:
$$P[X=4] = {5 \choose 1} \cdot \frac{5}{9} \cdot \frac{4 \choose 2}{8 \choose 4} = 5 \cdot \frac{5}{9} \cdot \frac{6}{70} = \frac{15}{63}$$
Indeed, $P[X=0] + P[X=2] + P[X=4] = 1$.
