Q.Prove that $^{n}C_{r} + \, ^{n}C_{r-1} = \, ^{n+1}C_{r}$, and verify it numerically for $n=6,\,r=3$.

This is Pascal's rule, proved in general by combining the two combination formulas over a common denominator, then confirmed with actual numbers.

General proof. Start from the definitions: $^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$ and $^{n}C_{r-1} = \dfrac{n!}{(r-1)!(n-r+1)!}$.

Factor $\dfrac{n!}{(r-1)!(n-r)!}$ out of both terms (since $r! = r(r-1)!$ and $(n-r+1)! = (n-r+1)(n-r)!$):

Regrouping the factorial parts: $\dfrac{n!\,(n+1)}{r(r-1)!\,(n-r+1)(n-r)!} = \dfrac{(n+1)!}{r!\,(n-r+1)!} = \,^{n+1}C_{r}$.

So $^{n}C_{r}+\,^{n}C_{r-1}=\,^{n+1}C_{r}$, as required.

Numerical verification for $n=6,\,r=3$: $^{6}C_{3} = \dfrac{6!}{3!3!} = 20$, and $^{6}C_{2} = \dfrac{6!}{2!4!}=15$. Sum $=20+15=35$. …

Losing track of which factorial term is common while combining the two fractions in the general proof — it helps to write $r!=r\cdot(r-1)!$ and $(n-r+1)!=(n-r+1)\cdot(n-r)!$ expli …