Bernoulli and binomial random variables Let $X_1, \cdots, X_n$ be iid Bernoulli, i.e. $X_i = 1$ with probability $p$, $X_i = 0$ with probability $1-p$. Now let $Y = \sum_{i=1}^{n}X_i$ be the number of "success" (i.e., when $X_i = 1$) among $n$ trials. Obviously $Y$ is binomial so $P(Y=y) = \binom{n}{y}p^y(1-p)^{n-y}$. 
However, what about the probability $P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n, Y=y)$? I know that this probability is $0$ if $\sum_{i=1}^{n}x_i \neq y$, but I am told that if $\sum_{i=1}^{n}x_i=y$ then $P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n, Y=y)= p^y(1-p)^{n-y}$ because "it is the probability of an independent sequence of $0$'s and $1$'s with a total of $y$ $1$'s and $(n-y)$ $0$'s." 
My question is why is the probability $p^y(1-p)^{n-y}$? In other words why don't we have the binomial coefficient ($\binom{n}{y}$) anymore? Is it because we already know the realized values of $X_1, X_2, \cdots, X_n$, so there is no need to account for the number ways the $0$'s and $1$'s are dispersed? 
 A: Yes, that is indeed the reason.
When we have a particular sequence of $(x_i)_{i=1}^n\in\{0,1\}^n$ where $~\sum_{i=1}^n x_i=y~$, then $\mathsf P(\{Y=y\}\cap \bigcap_{i=1}^n \{X_i=x_i\})$ is the probability of a particular sequence of $y$ success and $(n-y)$ failures.
So when $~\sum_{i=1}^n x_i=y~$ , then $~\mathsf P(\{Y=y\}\cap \bigcap_{i=1}^n \{X_i=x_i\}) ~{= \prod_{i=1}^n\mathsf P(X_i=x_i) \\= p^y (1-p)^{n-y}}$ 

Eg: for $n=4, p=1/3$ then  $\mathsf P(Y=3, X_1=1, X_2=1, X_3=0, X_4=1) ~{= pp(1-p)p \\ = p^3(1-p) \\ =\frac{2}{81}}$
A: As soon as $X_1=x_1,\dots,X_n=x_n$,  you will have $Y=\sum_{i=1}^{n}x_i$ a.e.
Then $P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n, Y=y)\\=P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n, Y=y|Y=\sum_{i=1}^{n}x_i)P(Y=\sum_{i=1}^{n}x_i)\\=P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n|Y=\sum_{i=1}^{n}x_i)\\=P(X_1 = x_1, X_2 = x_2, \cdots, X_n = x_n)=\prod_{i=1}^{n}P(X_i=x_i)\\=\prod_{i=1}^{n}p^{x_i}(1-p)^{1-x_i}=p^{\sum_{i=1}^{n}x_i}(1-p)^{n-\sum_{i=1}^{n}x_i}=p^y(1-p)^{n-y}$
otherwise $0$, since $P(Y\neq \sum_{i=1}^{n}x_i)=0$
