Q.Find the value of ${}^{15}C_4 + {}^{15}C_5$

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 $r$ given both ${}^nP_r$ and ${}^nC_r$' problem (since dividing one by the other isolates $r!$, hence $r$, immediately). Finally, ${}^nC_0+{}^nC_1+\cdots+{}^nC_n=2^n$ (the total number of subsets of an $n$-element set, including the empty set), and ${}^nC_r$ is greatest at the middle value $r=n/2$ when $n$ is even, or at either of the two equal middle values $r=\frac{n-1}{2}$ and $r=\frac{n+1}{2}$ when $n$ is odd — a fact used whenever a problem asks for the 'greatest value' of a binomial coefficient family.