Probability problem (withdraw balls from the urn) 
Problem:
  An urn contains 3 red and 7 black balls. Player A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn)

I have the solution:
$$\Bbb{P}(A)=\frac{3\cdot 9!+6\cdot 3 \cdot 7!+ 7\cdot6\cdot5\cdot4\cdot3\cdot5!+7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot3\cdot3!}{10!}$$
But i have no idea how they came up with this. Can someone explain  this? Thank you.

Well, the second entry on top is wrong (there should be another factor of $7$), so that could be the source of your confusion. Let's assume it's just a typo, and see how they might have gone about it.
There are $4$ mutually exclusive possibilities to consider for ways Player A can draw the red ball:

(i) The first ball drawn is red. Probability of this is $\dfrac3{10}$
(ii) The first two balls drawn are black and the third one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac38=\dfrac{7\cdot 6\cdot 3}{10\cdot 9\cdot 8}$.
(iii) The first four balls drawn are black and the fifth one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6}$.
(iv) The first six balls drawn are black and the seventh is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36\cdot\dfrac25\cdot\dfrac34=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$.

[Do you see why these probabilities are correct?]
Since these events are mutually exclusive, the probability that Player A gets the red ball is simply the sum of these probabilities. The least common denominator is clearly $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4=\frac{10!}{3!}$. It seems your book chose to use $10!$ as the common denominator, instead (probably to reduce the length of the answer). Rewriting the probabilities with $10!$ as denominator, we therefore have $$\begin{align}\Bbb P(A) &= \frac{3\cdot9!}{10!}+\frac{7\cdot 6\cdot 3\cdot 7!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}\\ &= \frac{3\cdot9!+7\cdot 6\cdot 3\cdot 7!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}.\end{align}$$
