Prove that the sum is sufficient using using the definition of sufficiency

If $$X_1,\ldots,X_n$$ is an IID random sample, with $$X_i\sim\,\text{Ber}(\theta)$$, prove that $$Y = \sum_i X_i$$ is sufficient using the definition of sufficiency (not the factorization criterion).

Now the definition of sufficiency I'm given is:

A statistic $$Y$$ is sufficient if the joint distribution of the sample given the statistic does not depend on $$\theta$$ or

$$P_\theta(X_1=x_1,\ldots, X_n=x_n | Y_n = y)= f(x_1,\ldots,x_n,y),$$ for some non-negative function $$f$$ not depending on $$\theta$$.

My attempt:

Since $$Y$$ is the sum of $$n$$ Bernoulli random variables, $$Y\sim \text{Bin}(n,\theta)$$. Hence

$$P_\theta(X_1=x_1,\ldots, X_n=x_n | Y_n = y) = \frac{P_\theta(X_1=x_1,\ldots, X_n=x_n, Y_n = y)}{{n \choose y} \theta^y (1-\theta)^{n-y}}$$

but here I'm stuck as I have no idea how to compute the probability in the numerator. Any help or hint is highly appreciated.

You are almost there. Note that after $$X_1,\ldots, X_n$$ has been fixed at $$x_1,\ldots, x_n$$, $$Y$$ is automatically fixed. Therefore, your numerator becomes
$$P_\theta(X_1=x_1,\ldots, X_n=x_n, x_1+\ldots+x_n = y) =\\=P_\theta(X_1=x_1,\ldots, X_n=x_n) \\=\theta^{\sum_i x_i} (1-\theta)^{n-\sum_i x_i}.$$