From Blitzstein & Hwang, Introduction to Probability p160-161:
Set-up
Blitzstein shows how to find the variance of the binomial distribution using indicator variables (the outcome of Bernoulli trials). I understand the reasoning.
Let's find the variance of $X \sim Bin(n, p)$ using indicator r.v.s to avoid tedious sums. Represent $X = I_1 + I_2 + … + I_n$, where $I_j$ is the indicator of the jth trial being a success. Each $I_j$ has variance $$Var(I_j)=E(I^2_j)-(E(I_j))^2=p-p^2=p(1-p)$$ (Recall that $I^2_j = I_j$, so $E(I^2_j) = E(I_j) = p$.)
Since the $I_j$ are independent, we can add their variances to get the variance of their sum: $$Var(X)=Var(I_1)+...Var(I_n)=np(1-p)$$
They proceed to find the variance via an alternate means (no pun intended).
Alternatively, we can find $E(X^2)$ by first finding $E\binom{X}{2}$. The latter sounds more complicated, but actually it is simpler since $\binom{X}{2}$ is the number of pairs of successful trials. Creating an indicator r.v. for each pair of trials, we have $$E\binom{X}{2}=\binom{n}{2}p^2\tag{1}$$
Thus, $$n(n-1)p^2=E(X(X-1))=E(X^2)-E(X)=E(X^2)-np\tag{2}$$ which again gives $$Var(X)=E(X^2)-(EX)^2=(n(n-1)p^2+np)-(np)^2=np(1-p)\tag{3}$$
Question
How are the authors justified in pursuing this line of reasoning. Where does $\binom{X}{2}$ come from? There is no mention of constraining on an even number of successful Bernoulli Trials so why would they count in pairs?
Part one is clear. $$ \left(\matrix{x \\ 2}\right) = \frac{x!\cdots}{2!(x-2)!} = \frac{x(x-1)}{2} = \frac{1}{2}x^2-\frac{1}{2}x $$ The pairs is useful relationship to get a solution for the expectation of the lhs.