Probability of 3 red balls out of 200 balls placed in 5 bins Given 3 red balls, 197 black balls, and 5 bins. 
The balls are randomly distributed with 40 balls in each bin. 
What is the probability that the three red balls are in three separate bins
(each of the three bins has one red ball;
i.e., no two red balls are in the same bin)?
Different variants of this question has been asked –
but somehow I still have trouble computing the correct probability.
One answer I found was:
$$\frac{\left(\begin{array}{l}
5 \\
1
\end{array}\right)\left(\begin{array}{l}
4 \\
1
\end{array}\right)\left(\begin{array}{l}
3 \\
1
\end{array}\right)}{125} = 12/25$$
It only focuses on red balls, and there are 125 ways that 3 red balls can be allocated in bins.
However, another alternative way of computing this seems to be:
$$
\frac{5 \times {3 \choose 1} {199 \choose 39} {3 \choose 1} {199 \choose 39} {3 \choose 1} {199 \choose 39} {200 \choose 40} {200 \choose 40}}{{200 \choose 40}{200 \choose 40}{200 \choose 40}{200 \choose 40}{200 \choose 40}} = 27/25
$$
Obviously it seems that the alternative way is incorrect (it's above 1 as a probability). So my question is: is this alternative way of computing incorrect? How so? How can I fix it?
 A: Imagine $200$ slots divided into $5$ groups of $40$. By far the simpler computation is just distributing the $3$ red balls one by one into different groups.
The first red ball can go anywhere, and  $Pr = \frac{160}{199}\frac{120}{198} = \frac{3200}{6567}$
A: Let's enumerate all of the balls as follows: $$R_1R_2R_3B_1B_2B_3\dots B_{196}B_{197}$$
Using this enumeration, we see that there is a one to one correspondence between placing the $200$ balls into the $5$ bins and two hundred digit numbers whose digits are sampled from $\{1,2,3,4,5\}$ with replacement. For example, the two hundred digit number $$143555\dots55$$ signifies that we put $R_1$ into bin $1$, $R_2$ into bin $4$, $R_3$ into bin  $3$, and the remaining $197$ black balls into the fifth bin.
The numbers of ways to distribute the $200$ balls into the $5$ bins in such a way that the the red balls all belong to different bins is precisely the number of two hundred digit numbers whose first three digits are all distinct. The corresponding probability is $$\frac{{5 \choose 3}3!\times 5^{197}}{5^{200}}=\frac{12}{25}$$
If we're distributing the balls in such a way that each bin contains $40$ balls, then this becomes $$\frac{{5 \choose 3}3!\times\frac{197!}{(39!)^3 (40!)^2}}{\frac{200!}{(40!)^5}}=\frac{3200}{6567}$$
A: The alternative way is attempting to calculate ways to place the 3 red balls and 39 others into three boxes, and 40 balls into the two other boxes, when selecting from the 200 balls ... each time.
The error is in not decrementing the source count for the balls.  The process is done without replacement.
$$\small{\dfrac{\dbinom 53\, \dbinom{3}{1}\dbinom{197}{39}\,\dbinom{2}{1}\dbinom{158}{39}\,\dbinom{1}{1}\dbinom{119}{39}\,\dbinom{0}{0}\dbinom{80}{40}\,\dbinom{0}{0}\dbinom{40}{40}}{\dbinom{200}{40}\dbinom{160}{40}\dbinom{120}{40}\dbinom{80}{40}\dbinom{40}{40}}\\=\dfrac{\dbinom 53\dbinom{3}{1,1,1,0,0}\dbinom{197}{39,39,39,40,40}}{\dbinom{200}{40,40,40,40,40}}\\=\dfrac{5!}{3!2!}\dfrac{3!}{1!^30!^2}\dfrac{197!}{39!^340!^2}\dfrac{40!^5}{200!}\\=\dfrac{3200}{6567}}$$

Another approach.  Count ways to place the three balls among five groups of forty places when selecting three among two hundred places.  Which produces the same result.
$$\dfrac{\dbinom 53~\dbinom{40}{1}\dbinom{40}1\dbinom{40}1\dbinom{40}0\dbinom{40}0}{\dbinom{200}3}=\dfrac{3200}{6567}$$
