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! = 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 × 7 × 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 = (3 × 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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