How do I show the following?
$$ \sum_{x=0}^{n} x {N_1 \choose {n-x}} {N_2 \choose x} = N_2 {N_1 + N_2 - 1 \choose n-1} $$
I tried breaking down the left hand side into factorials and pulling out $N_2$, but that did not help. How does one deal with these summmations in general?
$$ \binom{N_2}{x} = \frac{N_2}{x}\binom{N_2-1}{x-1} $$
With this, the sum gets transformed to
$$ \sum_{x=1}^n x\binom{N_1}{n-x}\binom{N_2}{x} = N_2\sum_{x=1}^n\binom{N_1}{n-x}\binom{N_2-1}{x-1}. $$
The rest is easy with a combinatorial argument. Starting the index with $0$ or $1$ doesn't make a difference.
There are $N_1$ men and $N_2$ women. You want to select a team of $n$ people; and from this team of $n$, select a leader who is a woman. How many ways can you do this?
(1) Pick a number $x$ between $1$ and $n$. Select $x$ women, then select a leader from these $x$, then select $n-x$ men. There are $$\sum_{x=1}^{n}{N_2\choose x}x{N_1\choose n-x}$$ ways to do this. This is the LHS of your identity (the $x=0$ term contributes nothing; if $x$ exceeds the number of women, then selecting $x$ women is impossible, but $N_2\choose x$ is zero as well.)
(2) Select a woman to be leader, then select $n-1$ people for the rest of the team. There are $$ N_2{N_1+N_2-1\choose n-1} $$ ways to do this. This is the RHS of your identity.
EDIT: You can generalize to the problem of selecting a team of $n$ people containing a subteam of $k$ women. The same argument gives: $$\sum_{x=k}^n{N_2\choose x}{x\choose k}{N_1\choose n-x}={N_2\choose k}{N_1+N_2-k\choose n-k}\tag1 $$
Just to complete the Nilabro Saha's answer we have $$ S=\sum_{k=0}^{n}k\dbinom{N_{1}}{n-k}\dbinom{N_{2}}{k}=N_{2}\sum_{k=0}^{n}\dbinom{N_{1}}{n-k}\dbinom{N_{2}-1}{k-1} $$ and using the Pascal's triangle $$S=N_{2}\sum_{k=0}^{n}\dbinom{N_{1}}{n-k}\dbinom{N_{2}}{k}-N_{2}\sum_{k=0}^{n}\dbinom{N_{1}}{n-k}\dbinom{N_{2}-1}{k} $$ and so using the Chu-Vandermonde identity we get $$S=N_{2}\left(\dbinom{N_{1}+N_{2}}{n}-\dbinom{N_{2}+N_{2}-1}{n}\right)=\color{red}{N_{2}\dbinom{N_{2}+N_{2}-1}{n-1}}$$ as wanted.
Here is a technique based upon the coefficient of operator $[t^q]$ to denote the coefficient of $t^q$ in a series. This way we can write e.g. \begin{align*} \binom{p}{q}=[t^q](1+t)^p \end{align*}
We obtain \begin{align*} \sum_{x\geq 1}&x \binom{N_1}{n-x}\binom{N_2}{x}\tag{1}\\ &=N_2\sum_{x\geq 1}\binom{N_1}{n-x}\binom{N_2-1}{x-1}\tag{2}\\ &=N_2\sum_{x\geq 0}\binom{N_1}{n-x-1}\binom{N_2-1}{x}\tag{3}\\ &=N_2\sum_{x\geq 0}[t^{n-x-1}](1+t)^{N_1}[u^x](1+u)^{N_2-1}\tag{4}\\ &=N_2[t^{n-1}](1+t)^{N_1}\sum_{x\geq 0}t^{x}[u^x](1+u)^{N_2-1}\tag{5}\\ &=N_2[t^{n-1}](1+t)^{N_1}(1+t)^{N_2-1}\tag{6}\\ &=N_2[t^{n-1}](1+t)^{N_1+N_2-1}\\ &=N_2\binom{N_1+N_2-1}{n-1}\tag{7} \end{align*} and the claim follows.
Comment:
In (1) we start the index from $x=1$ due to the factor $x$ and we increase the upper limit to $\infty$ without changing anything, since we are adding zeros only.
In (2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
In (3) we shift the index to start from $x=0$.
In (4) we apply the coefficient of operator twice.
In (5) we use the linearity of the coefficient of operator and use the rule \begin{align*} [t^{p-q}]A(t)=[t^p]t^qA(t) \end{align*}
In (6) we apply the substitution rule of the coefficient of operator with $u:=t$ \begin{align*} A(t)=\sum_{x=0}^\infty a_xt^x=\sum_{x=0}^\infty t^x[u^x]A(u) \end{align*}
In (7) we select the coefficient of $t^{n-1}$.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{x = 0}^{n}x{N_{1} \choose n - x}{N_{2} \choose x} & = \sum_{x = 0}^{\infty}x{N_{2} \choose x}\ \overbrace{% \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{N_{1}} \over z^{n - x + 1}} \,{\dd z \over 2\pi\ic}}^{\ds{N_{1} \choose n - x}} \\[5mm] & = \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{N_{1}} \over z^{n + 1}}\ \overbrace{% \sum_{x = 0}^{\infty}{N_{2} \choose x}x\,z^{x}} ^{\ds{N_{2}\,z\,\pars{1 + z}^{N_{2} - 1}}}\ \,{\dd z \over 2\pi\ic} \\[5mm] & = N_{2}\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{N_{1} + N_{2} - 1} \over z^{n}} \,{\dd z \over 2\pi\ic} = \bbx{\ds{N_{2}{N_{1} + N_{2} - 1 \choose n - 1}}} \end{align}