The value of $$'n'$$ for which $$^{n - 1}{C_4} - {\,^{n - 1}}{C_3} - \frac{5}{4} \cdot {\,^{n - 2}}{P_2} < 0,$$       where $$n \in N$$

Solution :
We have, $$^{n - 1}{C_4} - {\,^{n - 1}}{C_3} - \frac{5}{4} \cdot {\,^{n - 2}}{P_2} < 0$$
$$\eqalign{ & \Rightarrow \frac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{4!}} - \frac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{3!}} - \frac{5}{4}\left( {n - 2} \right)\left( {n - 3} \right) < 0 \cr & \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 11} \right)\left( {n + 2} \right) < 0 \cr & \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 11} \right) < 0\left[ {\because n + 2 > 0{\text{ for }}n \in N} \right] \cr & \Rightarrow n \in \left( { - \infty ,2} \right) \cup \left( {3,11} \right) \cr} $$

$$\eqalign{ & \Rightarrow n \in \left( {0,2} \right) \cup \left( {3,11} \right) \cr & \Rightarrow n = 1,4,5,6,7,8,9,10 \cr} $$
But $$^{n - 1}{C_4}$$  and $$^{n - 2}{P_2}$$  both are meaningful for $$n \geqslant 5.$$
Hence, $$n = 5,6,7,8,9,10.$$