About identical objects and distinct boxes kind of questions Problem 1 :Find the probability that exactly two boxes are empty when $5$ identical balls are randomly distributed to $5$ distinguishable boxes.
Solution : $C(5,2)C(2+3-1,2)/C(5+5-1,5)$
Problem 2 :Suppose that $200$ identical balls are randomly distributed into $100$ distinct boxes. Find the probability that there are exactly $27$ balls in the first $50$ boxes.
Solution : $C(200,27)(1/2)^{27}(1/2)^{73}$
Why we can't make the "stars and bars" argument for problem 2? Like, why the answer is not $C(200,27)C(50+27-1,27)C(50+183-1,183)/C(100+200-1,200)$ ?
Thanks in advance for your help.
 A: The problem is that the
$$\binom{200 + 100 - 1}{100 - 1} = \binom{200 + 100 - 1}{200}$$
ways to distribute $200$ identical balls into $100$ distinct boxes are not equally likely to occur.  There is only one way to place all $200$ balls in the first box.  However, there are 
$$\binom{200}{2}\binom{198}{2}\binom{196}{2} \ldots \binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways to distribute two balls to each of the $100$ distinct boxes.
The problem is solved by using the binomial distribution.  If an event has probability $p$ of occurring in each trial, it has probability $1 - p$ of not occurring.  The probability that it occurs exactly $k$ times in $n$ trials is 
$$P(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $\binom{n}{k}$ counts the number of orders in which an event can occur $k$ times in $n$ trials, $p^k$ is the probability the event occurs $k$ times, and $(1 - p)^{n - k}$ is the probability that the complementary occurs $n - k$ times.  
In our example, $n = 200$, $k = 27$, and $n = 1/2$ since placement in $50$ of the $100$ boxes results in the desired outcome. 
