If $5. {^{n} P}_{3} = 4. {^{n + 1} P}_{3}$ , then find the value of $n$ .

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Solution

Given
$5. {^{n} P}_{3} = 4. {^{n + 1} P}_{3}$
$3 n (n - 1) (n - 2) = 4 (n + 1) n (n - 1)$
$5 (n - 2) = \frac{4 (n + 1) n (n - 1)}{n (n - 1)}$
$5 (n - 2) = 4 (n + 1)$
$5 n - 10 = 4 n + 4$
$5 n - 4 n = 4 + 10$
$n = 14$

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n

^{n} P_{5}

= 42

^{n} P_{3}

n \geq 5.

n

_{n} P_{5} = 42 (_{n} P_{3}) : n > 4

^{n - 1} P_{3} :^{n} P_{4} = 1.4

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^{n} P_{3} :^{n - 1} P_{3} = 5 : 4

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

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