Question:
If 12Pr = 1320, then r =__________.
Solution:
Given ${ }^{12} P_{r}=1320$
i. e $\frac{12 !}{(12-r) !}=1320$
$\Rightarrow(12-r) !=\frac{12 !}{1320}$
$\Rightarrow(12-r) !=\frac{12 \times 11 \times 10 \times 9 !}{12 \times 11 \times 10}$
$\Rightarrow(12-r) !=9 !$
$\Rightarrow 12-r=9$
$\Rightarrow r=12-9$
i. e $r=3$