Math, asked by govindm573, 11 months ago

how we solve this.... factorial question​

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Answered by komalAgrawal43
0

Definition of Factorial

Let n be a positive integer. n factorial, written n!, is defined by

n! = 1 × 2 × 3 × ... (n - 1) × n

The special case when n = 0, 0 factorial is given by: 0! = 1

Question 1

Evaluate the following expressions:

4!

5! × 5!

3! × 0!

4! / 0!

6! / (2! × 4!)

Solution to Question 1

4! = 1 × 2 × 3 × 4 = 24

5! × 5! = (1 × 2 × 3 × 4 × 5) 2 = 120 2 = 14400

3! × 0! = (1 × 2 × 3) × 1 = 6

4! / 0! = (1 × 2 × 3 × 4) / 1 = 24

6! / (2! × 4!)

= (1 × 2 × 3 × 4 × 5 × 6) / [ (1 × 2 ) × ( 1 × 2 × 3 × 4) ]

= 15

Question 2

Simplify the following expressions:

(n + 2)! / n!

(2n + 2)! / 2n!

(n - 1)! / (n + 1)!

Solution to Question 2

Expand the factorials

(n + 2)! / n!

= [ 1 × 2 × ... × n × (n + 1) × (n + 2) ] / [ 1 × 2 × ...× n ]

and simplify to obtain

= (n + 1)(n + 2)

Expand the factorials

(2n + 2)! / 2n!

= [ 1 × 2 × 3...(2n) × (2n + 1) × (2n + 2) ] / [ 1 × 2 × 3...2n ]

Simplify

= (2n + 1) × (2n + 2)

Expand the factorials

(n - 1)! / (n + 1)!

= [ 1 × 2 × 3...(n - 1) ] / [ 1 × 2 × 3...(n - 1) × n × (n + 1) ]

Simplify

= 1 / [ n × (n + 1) ]

Exercises

a) Evaluate (10! / 5!) / 10

b) Simplify (n + 1)! / n!

Answers to above exercises

a) (10! / 5!) / 10 = 3024

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

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