Probability with replacement - binomial distribution and combinatorics with replacements seem to give different answers I have come across a question about probabilities with replacement, and I would like to take a combinatorics approach to the question; however, it has failed, and I would like to know if I am doing something wrong. So let me first explain the question:
There are five balls with different colors: red, green, blue, white, black. The question is what is the probability of drawing exactly two red balls with replacement.
I know the answer is:
$$\left(\frac{1}{5}\right)^2 \cdot \left(\frac{4}{5}\right)^3 \cdot {5 \choose 2} = \frac{128}{625}$$
since there is a $\frac{1}{5}$ chance of choosing a red ball and $\frac{4}{5}$ chance of not choosing a red ball. The order does not matter as well so multiply it by $5 \choose 2$. This explanation makes sense and it is in agreement with binomial distribution.
Now there is another approach as well, which is the one that gives me the wrong answer.
$$P(\text{exactly 2 red balls}) = \frac{\text{number of ways of drawing exactly 2 red balls from 5 draws}}{\text{total number of ways of drawing 5 balls with replacement}}$$
For finding the total number of ways of drawing 5 balls with replacement, we have 5 choices for each draw so we $5^5$ in total. Now to find the number of ways of drawing exactly $2$ red balls from $5$ draws, I think we can use the formula for drawing with replacement which is given by
$$n+k-1 \choose k$$
So we can choose $2$ red balls in ${5+2-1 \choose 2} = {6 \choose 2}$ and the remaining $3$ balls could be chosen in ${5+3-1 \choose 3} = {7 \choose 3}$. When we put everything together, we will get
$$\frac{{6 \choose 2}\cdot {7 \choose 3}}{5^5} = \frac{21}{125}$$
We can see the two answers do not match. So my question is what went wrong in my approach? Have I calculated number of ways of drawing exactly $2$ red balls from $5$ draws wrongly?
 A: You should probably revisit where you found the $\binom{n+k-1}{k}$ formula and what exactly it is counting; I suspect you are misusing it.
In the second approach the numerator should be $\binom{5}{2}$ (choose which two of the five draws were red) times $4^3$ (choice of color for the the non-red draws) which matches the first approach.
A: The formula
$$\binom{n + k - 1}{k}$$
counts the number of ways of selecting $k$ objects selected from $n$ types of objects with repetition.  However, these selections are not equally likely to occur.  There is only one way to select a red ball on each of the five draws.  However, there are $5! = 120$ ways to select one red, one green, one blue, one white, and one black ball.
As angryavian has already pointed out, there are $\binom{5}{2}$ ways to select the two positions for the red balls and each of the remaining positions can be filled in four ways, so the number of favorable cases is
$$\binom{5}{2}4^3$$
yielding the probability of selecting exactly two reds
$$\Pr(\text{exactly two reds}) = \frac{\dbinom{5}{2}4^3}{5^5} = \frac{128}{625}$$
when five balls are selected with replacement from the five balls of different colors, which, as angryavian has pointed out, agrees with the answer you obtained using the Binomial distribution.
