Question

If ¹⁶C_r = ¹⁶C_{r + 2}, find ^rC₄.

Answer 1

Given:
${}^{16}{C}_r={}^{16}{C}_{r+2}$

$16=r+r+2$ [∵ Property 5: ${}^n{C}_x={}^n{C}_y\Rightarrow x=y$ or $x+y=n$]
$$\begin{gathered} \Rightarrow 2r+2=16 \\ \Rightarrow 2r=14 \\ \Rightarrow r=7 \end{gathered}$$

Now,${}^r{C}_4={}^7{C}_4$
$\Rightarrow {}^7{C}_4={}^7{C}_3$ [∵ ${}^n{C}_r{=}^n{C}_{n-r}$]
${\Rightarrow }^7{C}_4={}^7{C}_3=\frac{7}{3}\times \frac{6}{2}\times \frac{5}{1}{\times }^4{C}_0$ [∵ ${}^n{C}_r=\frac{n}{r}.{}^{n-1}{C}_{r-1}$ ]
${\Rightarrow }^7{C}_4=35$ [∵ ${}^n{C}_0=1$]