FACTORIALS

Factorial of a natural number n is the product of the first n natural numbers. It is denoted by n! :

That is,

n!  =  1 × 2 × 3 × 4 × ..............× (n − 1) × n

We read this symbol as “n factorial” or “factorial of n”. The notation n! was introduced by the French mathematician Christian Kramp in the year 1808.

Note that for a positive integer n

n!  =  n × (n - 1) × (n - 2) × ..............× 3 × 2 × 1

=  n(n - 1)! for n > 1

=  n(n - 1)(n - 2)! for n > 2

=  n(n - 1)(n - 2)(n - 3)! for n > 3 and so on

Observe that, 

1!  =  1

2!  =  2 × 1  =  2

3!  =  3 × 2 × 1  =  6

4!  =  4 × 3 × 2 × 1  =  24

5!  =  5 × 4 × 3 × 2 × 1  =  720

.......  =  .....................

15!  =  15 × 14 × 13 × .......... 3 × 2 × 1  =  1307674368000

We will require zero factorial in the latter sections of this chapter. It does not make any sense to define it as the product of the integers from 1 to zero. So,

we define 0! = 1.

Note that 0! = 1 is evident by substituting n = 0 in the equation

(n + 1)!  =  (n + 1) × n!

(0 + 1)!  =  (0 + 1) × 0!

1!  =  × 0!

1!/1  =  0!

1  =  0!

This way, we talk of factorial for non-negative integers. Note that factorials can be extended to certain negative numbers and also to complex numbers. 

Example 1 :

Find the value of 

8!

Solution :

8!  =  8 × × 6 × 5 × 4 × 3 × 2 × 1 

=  40320  

Example 2 :

Simplify :

10!/7!

Solution :

10!/7!  =  (10 × 9 × 8 × 7!) / 7!

=  10 × 9 × 8

=  720

Example 3 :

Find the value of 

(3!)2/9

Solution :

(3!)2/9  =  (× 2 × 1)2 / 9

  =  9 × 4 × 1 / 9

  =  4 × 1

=  4

Example 4 :

Simplify :

(3! × 4!)/5!

Solution :

(3! × 4!)/5!  =  (3! × 4!/ (5 × 4!)

=  3!/5

=   (3 × 2 × 1) / 5

  =  6/5

Example 5 :

Simplify :

9!/(5× 3!)

Solution :

9!/(5! × 3!)  =  (9 × 8 × 7 × 6 × 5!) / (5! × 3!)

=  (9 × 8 × 7 × 6) / 3!

=  (9 × 8 × 7 × 6) / (3 × 2 × 1)

=  (9 × 8 × 7 × 6) / 6

=  9 × 8 × 7

=  504

Example 6 :

Evaluate n! / [r!(n - r)!] when n = 7 and r = 5.  

Solution :

n! / [r!(n - r)!]  =  7! / [5!(7 - 5)!]

=  7! / (5! × 2!)

=  (7 × 6 × 5!) / (5! × 2!)

=  (7 × 6) / 2

=  7 × 3

=  21

Example 7 :

Evaluate n! / [r!(n - r)!] for any n with r = 3.  

Solution :

n! / [r!(n - r)!]  =  n! / [3!(n - 3)!]

=  n(n - 1)(n - 2)(n - 3)! / [3!(n - 3)!]

=  n(n - 1)(n - 2) / 3!

=  n(n - 1)(n - 2) / (3 × 2 × 1)

=  n(n - 1)(n - 2) / 6

Example 8 :

If 6  =  6!/n!, then, find the value of n.  

Solution :

6  =  6!/n!

Multiply each side by n!. 

n! × 6  =  6!

Divide each side by 6.

n!  =  6!/6

n!  =  (1 × 2 × 3 × 4 × 5 × 6) / 6

n!  =  1 × 2 × 3 × 4 × 5

n!  =  5!

n  =  5

Example 9 :

If n! + (n - 1)!  =  30, then, find the value of n.  

Solution :

n! + (n - 1)!  =  30

n(n - 1)! + (n - 1)!  =  30

(n - 1)! × (n + 1)  =  30

(n - 1)! × (n + 1)  =  6 × 5

(n - 1)! × (n + 1)  =  (3 × 2 × 1) × 5

(n - 1)! × (n + 1)  =  3! × 5

Equate (n - 1)! to 3!. 

(n - 1)!  =  3!

n - 1  =  3

n  =  4

Example 10 :

What is the unit digit of the sum 2! + 3! + 4! + ........ + 22!?

Solution :

2!  =  2 × 1  =  2

3!  =  3 × 2 × 1  =  6

4!  =  4 × 3 × 2 × 1  =  24

 5!  =  5 × 4!  =  5 × 24  =  120

6!  =  6 × 5!  =  6 × 120  =  720

7!  =  7 × 6!  =  7 × 720  =  5040

8!  =  8 × 7!  =  8 × 5040  =  40320

From 5! onwards for all n!, the unit digit is zero and hence the contribution to the unit digit is through (2! + 3! + 4!) only.  

That is, 

2! + 3! + 4!  =  2 + 6 + 24

=  32

Therefore the required unit digit is 2.

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