Question

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$\mathsf {The \: value \: of \: ( {(n!)}^{n}) \: if \: (n) + (n - 1) + (n - 2) = n(n - 1)(n - 2)}$

$\mathsf {where \: {n}^{3} > 9, \: a \: positive \: number \: is}$


1) 316


2) 116


3) 216


4) 106



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Answer 1

Answer:

Option(3)

Step-by-step explanation:

Given Equation is n + (n - 1) + (n - 2) = n(n - 1)(n - 2)

⇒ n + n - 1 + n - 2 = n[n² - 3n + 2]

⇒ 3n - 3 = n³ - 3n² + 2n

⇒ n³ - 3n² - n + 3 = 0

⇒ n²(n - 3) - (n - 3) = 0

⇒ (n - 3)(n² - 1) = 0

⇒ (n - 3)(n + 1)(n - 1) = 0

⇒ n = 3, -1, 1

Given that the value of n³ > 9.

So, the value of n should be 3. {3³ = 27 > 9}

Now,

Given: ((n!)ⁿ)

= (3!)³

= 6³

= 216

Therefore, the value of ((n!)ⁿ = 216.

Hope it helps!

Answer 2

Answer:

Please see the attachment