Question

If ²ⁿC₃ : ⁿC₂ = 44 : 3, find n.

Answer 1

Given: 

\[2 n_{C_3} : n_{C_2} = 44: 3\]

\[\frac{2 n_{C_3}}{n_{C_2}} = \frac{44}{3}\]
\[ \Rightarrow \frac{2n!}{3! (2n - 3)!} \times \frac{2! (n - 2)!}{n!} = \frac{44}{3}\]
\[ \Rightarrow \frac{2n (2n - 1) (2n - 2)}{3 n (n - 1)} = \frac{44}{3}\]
\[ \Rightarrow (2n - 1) (2n - 2) = 22 (n - 1)\]
\[ \Rightarrow 4 n^2 - 6n + 2 = 22n - 22\]
\[ \Rightarrow 4 n^2 - 28n + 24 = 0\]
\[ \Rightarrow n^2 - 7n + 6 = 0\]
\[ \Rightarrow n^2 - 6n - n + 6 = 0\]
\[ \Rightarrow n (n - 6) - 1(n - 6) = 0\]
\[ \Rightarrow (n - 1) (n - 6) = 0\]

\[\Rightarrow n = 1\]   or,  

\[n = 6\]

Now, 

\[n = 1 \Rightarrow 2_{C_3} : 2_{C_2} = 44: 3\]

But, this is not possible.

∴ \[n = 6\]