SIMPLIFYING FACTORIALS WITH VARIABLES

Simplify the following expressions.     

Problem 1 :

(n + 2)! / n!

Solution :

= (n + 2)! / n!

Writing (n + 2)! in descending order, we get

= ((n + 2) ∙ (n + 1) ∙ n!) / n!

= (n + 2) (n + 1)

Problem 2 :

n / n!

Solution :

= n / n!

= n / n(n - 1)! 

Cancelling n in both numerator, we get

= 1 / (n - 1)!

Problem 3 :

(n - 1)! ∙ n! / (n!)²

Solution :

= ((n - 1)! ∙ n!) / (n!)²

= ((n - 1)! ∙ n!) / (n! ∙ n!)

= (n - 1)! / n!

= (n - 1)! / n(n - 1)!

= 1/n

Problem 4 :

(n + 5)! / (n + 1)!

Solution :

= (n + 5)! / (n + 1)!

= (n + 5) (n + 4) (n + 3) (n + 2) (n + 1)! / (n + 1)!

= (n + 5) (n + 4) (n + 3) (n + 2)

Problem 5 :

((n + 1)!)³ / (n!)³

Solution :

= ((n + 1)!)³ / (n!)³

= [(n + 1) (n!)]3 / (n!)³

= [(n + 1)³ (n!)³] / (n!)³ 

= (n + 1)³

Problem 6 :

(n² - 1)! / (n²)!

Solution :

= (n² - 1)! / (n²)!

= (n² - 1)! / n² (n² - 1)!

= 1/n²

Problem 7 :

(2n)! / (2n - 2)! ∙ 2!

Solution :

= (2n)! / (2n - 2)! ∙ 2!

= (2n) (2n - 1) (2n - 2)!  / (2n - 2)!. 2!

= 2n (2n - 1) / (2 ∙ 1)           

= n (2n - 1)

Problem 8 :

(n - 1)! (n² + n)

Solution :

= (n - 1)! (n² + n)

= (n - 1)! n(n + 1)

Considering n+1 and n,

• when we reduce 1 from n + 1, we will get n.
• when we reduce 1 from n, we will get n - 1.

So, simply we can write it as (n + 1)!

(n - 1)! n(n + 1) = (n + 1)!

Problem 9 :

(n + 5) (n + 4) (n + 3)!

Solution :

Considering n + 5 and n + 4,

• when we reduce 1 from n + 5, we will get n + 4
• When we reduce 1 from n + 4, we will get n + 3.

So, simply we can write it as (n + 5)!.

(n + 5) (n + 4) (n + 3)! = (n + 5)!

Problem 10 :

n! (n² + 3n + 2)

Solution :

Since (n² + 3n + 2) is a quadratic polynomial, we can find the factors of this quadratic polynomial.

n² + 3n + 2 = (n + 1)(n + 2)

n! (n² + 3n + 2) = n!(n + 1)(n + 2)

Considering (n + 2) and (n + 1), 

• By reducing 1 from n + 2, we will get n + 1.
• By reducing 1 from n + 1, we will get n.

So, n!(n + 1)(n + 2) = (n + 2)!