If (n + 1)! = 12 × [(n – 1)!], find the value of n.

1.	If (n + 1)! = 12 × [(n – 1)!], find the value of n.
Answer» To Find: Value of n Given: (n+1)! = 12× [(n-1)!] Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1 Now, (n+1)! = 12× [(n-1)!] ⇒ (n+1) × (n) × [(n-1)!] = 12 × [(n-1)!] ⇒ (n+1) × (n) = 12 ⇒ n²+n = 12 ⇒ n²+n-12 = 0 ⇒ (n-3) (n+4) = 0 ⇒ n = 3 or, n = -4 But, n=-4 is not possible because in case of factorial (!) n cannot be negative. Hence, n=3 is the correct answer.

Answer»

To Find: Value of n

Given: (n+1)! = 12× [(n-1)!]

Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1

Now, (n+1)! = 12× [(n-1)!]

⇒ (n+1) × (n) × [(n-1)!] = 12 × [(n-1)!]

⇒ (n+1) × (n) = 12

⇒ n²+n = 12

⇒ n²+n-12 = 0

⇒ (n-3) (n+4) = 0

⇒ n = 3 or, n = -4

But, n=-4 is not possible because in case of factorial (!) n cannot be negative. Hence, n=3 is the correct answer.

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