Question
Evaluate the following:
$^{10}\text{P}_3$

Answer

We have,

$^{10}\text{P}_4 = \frac{10!}{(10-4)!}\Big[\because\ ^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ 

$=\frac{10!}{6!}$

$= \frac{10\times9\times8\times7 \times6!}{6!}$

$= 5040$

Hence, $^{10}\text{P}_4= 5040$

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