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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.

$$ \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$

ProgC
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    This is called Pascal's Rule. See also the answers here for combinatorial applications. – Dietrich Burde Sep 29 '18 at 16:58
  • I wrote about the Inclusion-Exclusion Principle and Bonferroni Inequality in here. See Lemma 1 for the application of the Pascal's Triangle. – Sungjin Kim Sep 29 '18 at 19:14
  • Here's one called as Hockey-Stick Identity: https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1?rq=1 – Sungjin Kim Sep 29 '18 at 19:19
  • Here I found one more : https://math.stackexchange.com/questions/1095785/binomial-coefficients-identity-sum-i-binomn-ik-1-binomn1k1?rq=1 – Sungjin Kim Sep 29 '18 at 19:34

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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 $n=1$, $$ \sum_{d|n}\lambda_d^+ = \lambda_1^+ = \mu(1)=1. $$

Sungjin Kim
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