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 \[x + y = n\]

\[\Rightarrow 2r + 2 = 16\]
\[ \Rightarrow 2r = 14\]
\[ \Rightarrow r = 7\]

Now,

\[{}^r C_4 = {}^7 C_4\]

\[\Rightarrow {}^7 C_4 = {}^7 C_3\] [∵\[{}^n C_r =^n C_{n - r}\]

\[\Rightarrow^7 C_4 = {}^7 C_3 = \frac{7}{3} \times \frac{6}{2} \times \frac{5}{1} \times^4 C_0\]
[∵\[{}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}\]]

\[\Rightarrow^7 C_4 = 35\] [∵\[{}^n C_0 = 1\]]