Question

If ⁿC₁₀ = ⁿC₁₂, find ²³C_n.

Answer 1

Given: ⁿC₁₀ = ⁿC₁₂
We have,

\[{}^n C_{10} = {}^n C_{12}\]

\[\Rightarrow n = 12 + 10 = 22\]  [∵\[{}^n C_x = {}^n C_y \Rightarrow x = y\]]  or, 

\[n = x + y\]

Now,

\[{}^{23} C_{22} = {}^{23} C_1\] [∵ \[{}^n C_r = {}^n C_{n - r}\]]

\[\Rightarrow {}^{23} C_{22} =^{23} C_1 = \frac{23}{1} \times^{22} C_0\] [∵\[{}^n C_r = \frac{n}{r} {}^{n - 1} C_{r - 1}\]]

\[\Rightarrow^{23} C_{22} = 23\] [∵\[{}^n C_0 = 1\]]