Find the value of n if $^{n} P_{4} = 5 (^{n} P_{3})$ .

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Solution

$^{n} P_{4} = 5 (^{n} P_{3})$
$⟹ \frac{n!}{(n - 4)!} = 5 \times \frac{n!}{(n - 3)!}$
$⟹ 1 = \frac{5}{n - 3}$
$⟹ n - 3 = 5$
$⟹ n = 8$

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^{n} p_{4} = 20 \times^{n} p_{2}

nP4=20(nP2) then find the value of n

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^{n - 1} P_{3} :^{n} P_{4} = 1 : 9.

^{n} P_{4} = 20^{n} P_{2}

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