Evaluate each of the following :
(i) $8_{P_{3}}$ (ii) $10_{P_{4}}$
(iii) $6_{P_{6}}$ (iv) P(6,4)

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Solution

(i) $8_{p_{3}}$
We have,
$8_{p_{3}} = \frac{8!}{(8 - 3)!} [∵ n_{p_{r}} = \frac{n!}{(n - r)!}]$
= $\frac{8 \times 7 \times 6 \times 5!}{5!}$
= 336
Hence, $8_{P_{3}} = 336$
We have,
$10_{P_{4}} = \frac{10!}{(10 - 4)!} [∵ n_{p_{r}} = \frac{n!}{(n - r)!}]$
= $\frac{10!}{6!}$
= $\frac{10 \times 9 \times 8 \times 7 \times 6!}{6!}$
= 5040
$∴ 10_{P_{4}} = 5040$
(iii) $6_{P_{6}}$
We have,
$6_{P_{6}} = \frac{6!}{(6 - 6)!} [∵ n_{P_{r}} = \frac{n!}{(n - r)!}]$
= $\frac{6!}{0!}$
= $\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{1} [∵ 0! = 1]$
= 720
Hence, $6_{p_{6}} = 720$ .
(iv)P(6,4)
We have,
P(6,4) = $\frac{6!}{(6 - 4)!} [∵ n_{p_{r}} = \frac{n!}{(n - r)!}]$
= $\frac{6!}{2!}$
= $\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2!}$
= 360
Hence, P(6,4)= 360.

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