A box contains $10$ red balls and $10$ blue balls.
What is the probability of choosing exactly $5$ red balls and $5$ blue balls if 10 balls are chosen randomly?
This is my answer:
How many ways to arrange $5B5R$ — $10!/5!5!$
Each arrangement have probability of $10/20*9/19*8/18*7/17*6/16*10/15*9/14*8/13*7/12*6/11 = (10!/5!)^2/(20!/10!) = (10!)^3 / ((5!)^2(20!))$
^ MULTIPLY WITH $10!/5!5! = (10!)^4/((5!)^4*20!)$
Can someone confirm that this is correct
Best Answer
It is easier to explain and do the calculations by temporarily assuming that the balls are all labeled.
There are $\binom{20}{10}$ number of ways to pick ten balls. There are $\binom{10}{5}\cdot \binom{10}{5}$ ways to pick five red and five blue balls.
This gives the probability then as:
$$\frac{\binom{10}{5}^2}{\binom{20}{10}}$$
If you insist on writing with factorials, then this expands as
$$\frac{\binom{10}{5}^2}{\binom{20}{10}}=\frac{10!^2}{5!^4}\cdot\frac{10!^2}{20!}$$ which matches your attempt. I would personally prefer to see the answer as I first gave it however as it makes it much more obvious what the logic behind the answer was and it can be verified to be correct at a glance.