An application of this binomical identity $ \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1} $ I need a clarification on this manner, in terms of "When do I apply this form of 
identity". I managed to proove it algebraically and combinatorially, using 
"Pascal's triangle". so a proof is not needed. However I am not comprehending what it's 
application in terms of combinatoric(or any) exersice.
$$ \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
 A: As I commented, it is most useful in proving the identity for $n\geq 1$, 
$$
\sum_{j=0}^m(-1)^j \binom nj = (-1)^m \binom {n-1}m.
$$
Then this identity leads to the proof of Bonferroni inequalities. 
This identity is also useful in Brun's Sieve methods. The following set-up is outlined in Montgomery & Vaughan's book 'Multiplicative Number Theory I'. Let $S(x,y;P)$ be the number of integers $n$ in the interval $x<n\leq x+y$ which are coprime to $P$. To find an upper bound of $S(x,y;P)$, we seek for $\lambda_d^+$ such that 
$$
\sum_{d|n}\lambda_d^+\geq \begin{cases} 1 &\mbox{ if } n=1 \\ 0 &\mbox{ otherwise.}\end{cases}
$$
Brun's sieve method, the $\lambda_d^+$ is determined by the following:
$$
\lambda_d^+ = \begin{cases} \mu(d) &\mbox{ if } w(d)\leq 2r \\ 0 &\mbox{ otherwise,}\end{cases}
$$
where $w(d)$ is the number of distinct prime factors of $d$, and $w(1)=0$.
Then for $n>1$, 
$$
\begin{align}
\sum_{d|n}\lambda_d^+ &= \sum_{j=0}^{2r} \sum_{d|n, w(d)=j}\mu(d)\\
&=\sum_{j=0}^{2r} (-1)^j \binom{w(n)}{j} = (-1)^{2r} \binom{w(n)-1}{2r}\\
&=\binom{w(n)-1}{2r}\geq 0.
\end{align}
$$
and for $n=1$, 
$$
\sum_{d|n}\lambda_d^+ = \lambda_1^+ = \mu(1)=1.
$$
