Permutation and Combinations — MATHS STD 11 Science — Question

2. n!

Solution:

We know that $\text{P}(\text{n}, \text{r}) = \ ^\text{n}\text{P}_\text{r} = \frac{\text{n}!}{(\text{n}-\text{r})!}$

Hence, $\text{P}(\text{n}, \text{r}) = \ ^\text{n}\text{P}_\text{n-1} = \frac{\text{n}!}{\big[\text{n}-\text{r}\big]!}$

$\text{P}(\text{n, n-1}) = \frac{\text{n!}}{(\text{n}-\text{n}+1)!} = \frac{\text{n}!}{1!} = \text{n}!$

Therefore, the value of $\text{P}(\text{n}, \text{n}-1)$ is $\text{n}!.$