Q.Show that $\dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r-1)!(n-r+1)!} = \dfrac{(n+1)!}{r!(n-r+1)!}$

Beyond the defining formula ${}^nC_r=\dfrac{n!}{r!(n-r)!}$, a handful of algebraic properties of combinations make many otherwise-heavy computations short and elegant, and this chapter uses them repeatedly. The complement property, ${}^nC_r={}^nC_{n-r}$, reflects the fact that choosing which $r$ objects to INCLUDE is equivalent to choosing which $n-r$ objects to LEAVE OUT — so it is always worth switching to whichever of $r$ or $n-r$ is smaller before expanding a computation by hand, and it is also the tool behind solving equations of the form ${}^nC_x={}^nC_y$, whose only two possibilities are $x=y$ or $x+y=n$. Pascal's Rule, ${}^nC_r+{}^nC_{r-1}={}^{n+1}C_r$, expresses how a triangle of combinations builds upward, and is proved by combining the two combination-fractions over a common denominator; it is invaluable for telescoping a running SUM of adjacent combinations down to a single binomial coefficient, and — read in reverse — for collapsing a DIFFERENCE of combinations sharing the same lower index. The identity ${}^nC_r = \dfrac{{}^nP_r}{r!}$ directly ties combinations back to permutations, and is often the fastest route through a 'find $n$ and $r$ give …