Question

Find the value of n such that :
nP5 =42nP3 , n > 4

Answer 1

$\huge\bold\red{Solution}$

nP5 = 42 nP3

n(n-1) (n-2) (n-3) (n-4) = 42 n(n-1) (n-2)

(n-3) (n-4) = 42

n² - 4n - 3n + 12 = 42

n² - 7n - 30 = 0

• By splitting middle term :

n² - 10n + 3n - 30 = 0

n(n-10) + 3(n-10) = 0

(n+3) (n-10) = 0

n =10 , -3 but n cannot be negative.

n = 10

• Thus the value of n is 10

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Answer 2

$\bold{Answer:}$

$\bold{10}$

$\bold{Step}\ - \ \bold{by}\ - \ \bold{step\ explanation:}$

$^nP_5 = 42 \ \times ^nP_3 \\ \\ \frac{n!}{(n - 5)!} = 42 \times \frac{n!}{(n - 3)!} \\ \\ \frac{n!}{(n - 5)!} \div \frac{n!}{(n - 3)!} = 42 \\ \\ \frac{n!}{(n - 5)!} \times \frac{(n - 3)!}{n!} = 42 \\ \\ \frac{(n - 3)!}{(n - 5)!} = 42 \\ \\ \frac{(n - 3)(n - 4)(n - 5)(n - 6)...... \times 2 \times 1}{(n - 5)(n - 6)...... \times 2 \times 1} = 42 \\ \\ (n - 3)(n - 4) = 42 \\ \\ n^2 - 7n + 12 = 42$

$n^2 - 7n + 12 - 42 = 0 \\ \\ n^2 - 7n - 30 = 0 \\ \\ n^2 + 3n - 10n - 30 = 0 \\ \\ n(n + 3) - 10(n + 3) = 0 \\ \\ (n - 10)(n + 3) = 0 \\ \\ \\ \therefore\ n = 10 \ \ \ \ \ ; \ \ \ \ \ n = -3 \\ \\ As \ n > 4,\ \ n \neq -3 \\ \\ \therefore\ \bold{n = 10}$

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