Evaluate each of the following:

Solution:

(i) ${ }^{8} P_{3}$

${ }^{n} P_{r}=\frac{n !}{(n-r) !}$

$\therefore{ }^{8} P_{3}=\frac{8 !}{(8-3) !}$

$=\frac{8 !}{5 !}$

$=\frac{8(7)(6)(5 !)}{5 !}$

$=8 \times 7 \times 6$

$=336$

(ii) ${ }^{10} P_{4}=\frac{10 !}{(10-4) !}$

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

$=\frac{10(9)(8)(7)(6 !)}{6 !}$

$=10 \times 9 \times 8 \times 7$

$=5040$

(iii) ${ }^{6} P_{6}=\frac{6 !}{(6-6) !}$

$=\frac{6 !}{0 !}$

$=\frac{6 !}{1}$      (Since, $0 !=1$ )

$=720$

(iv) $P(6,4)$

It can also be written as ${ }^{6} P_{4}$.

${ }^{6} P_{4}=\frac{6 !}{2 !}$

$=\frac{6(5)(4)(3)(2 !)}{2 !}$

$=6 \times 5 \times 4 \times 3$

$=360$