Should we draw a triangle?

By Pascal's identity, we have:

$\begin{aligned} ^{n-1}C_{k-1} + ^{n-1}C_k & = ^{n}C_{k} \\ \implies ^{14}C_{4} + ^{14}C_5 & = ^{15}C_5 \\ & = \frac {15!} {5!10!} \\ & = \frac {6\cdot 15\cdot 14!} {9\cdot 10\cdot 6!8!} \\ & = \frac {14!} {6!8!} \\ & = \boxed {^{14}C_6} \end{aligned}$

By the given notation $\begin{array}{rl} ^{14}C_4 + ^{14}C_5 &= ^{14}C_6\\ \dfrac{14!}{4!(14 - 4)!} + \dfrac{14!}{5!(14 - 5)!} &= \dfrac{14!}{6!(14 - 6)!}\\ \dfrac{14!}{4!10!} + \dfrac{14!}{5!9!} &= \dfrac{14!}{6!8!} \end{array}$ Dividing both sides by $14!$ gives $\dfrac{1}{4!10!} + \dfrac{1}{5!9!} = \dfrac{1}{6!8!}$ Thus, by multiplying both sides by $6!10!$ , $\begin{array}{rl} \dfrac{1}{4!10!} \cdot 6!10! + \dfrac{1}{5!9!} \cdot 6!10! &= \dfrac{1}{6!8!} \cdot 6!10!\\ \dfrac{6!}{4!} + \dfrac{6!10!}{5!9!} &= \dfrac{10!}{8!}\\ \dfrac{6 \cdot 5 \cdot 4!}{4!} + \dfrac{\left(6 \cdot 5!\right)\left(10 \cdot 9!\right)}{5!9!} &= \dfrac{10 \cdot 9 \cdot 8!}{8!}\\ 30 + 60 &= 90 \end{array}$ which gives $\boxed{\text{True}}$ .