Question

Evaluate each of the following:
(i) ${}^{8}P_3$

(ii) ${}^{10}P_4$

(iii) ${}^{6}P_6$

(iv) P(6, 4)

Answer 1

(i) ⁸P₃
ⁿP_r = $\frac{n!}{(n-r)!}$

∴ ⁸P₃ $=\frac{8!}{(8-3)!}$

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

$$\begin{gathered} =\frac{8\left(7\right)\left(6\right)(5!)}{5!} \\ =8\times 7\times 6 \\ =336 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){}^{10}{P}_4=\frac{10!}{(10-4)!} \\ =\frac{10!}{6!} \\ =\frac{10\left(9\right)\left(8\right)\left(7\right)(6!)}{6!} \\ =10\times 9\times 8\times 7 \\ =5040 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right){}^6{P}_6=\frac{6!}{(6-6)!} \\ =\frac{6!}{0!} \\ =\frac{6!}{1}(Since\; ,\; 0!\; =\; 1) \\ =720 \end{gathered}$$

(iv) P(6,4)
It can also be written as ⁶P₄ .
$$\begin{gathered} {}^6{P}_4=\frac{6!}{2!} \\ =\frac{6\left(5\right)\left(4\right)\left(3\right)(2!)}{2!} \\ =6\times 5\times 4\times 3 \\ =360 \end{gathered}$$