Factorial Notation

The continued product of first n natural numbers is called n factorial or factorial n.
It is denoted by | n or n!

Thus, | n or n! = 1. 2. 3. 4. .... (n -1) . n
                       = n (n -1)(n -2). .... 3.2.1          (in reverse order)

Meaning of zero factorial

According to the above definition, makes no sense. However we define 0! = 1

Note. When n is a negative integer or a fraction, n factorial is not defined.

Illustrative Examples

Example

Find the value of

  1. 25! / 23!
  2. 9! / [6!.3!]
  3. (7 -4)!
  4. 5! -4!
  5. 2! +3!

Solution

  1. 25!/23! = 25. 24 23!/ 23! = 25. 24 = 600
  2. 9! /[6! 3!] = 9.8. 7. 6! /[6! 3!] = 9. 8. 7/3! = 84
  3. (7 -4)! = 3! = 3. 2. 1 = 6
  4. 5! - 4! = (5. 4. 3. 2. 1) - (4. 3. 2. 1) = 120 -24 = 96
  5. 2! + 3! = (2. 1) + (3 . 2. 1) = 2 +6 = 8

Example

Find n, if n! /[2! (n -2)!] and n! /[4! (n -4)!] are in the ratio 2 : 1.

Solution

Given n! /[2! (n -2)!] : n!/[4!(n -4)!] = 2:1
=> [n! / 2! (| n -2)!].[4! (n -4)! /n!] = 2/1
=> 4.3.2!.(n -4)! /[2!.(n -2).(n -3).(n -4)!] = 2/1
=> 4.3 /[(n -2)(n -3)] = 2/1
=> (n -2)(n -3) = 6
=> n² -5 n = 0      =>    n (n -5) = 0
=>   n = 0 or n = 5
But, for n = 0, (n -2)! and (n -4)! are not meaningful, therefore, n = 5.

Example

Prove that 33! is divisible by 216. What is the largest integer n such that 33! is divisible by 2n?

Solution

33! = 1. 2. 3. 4. 5. 6. ... . 29. 30. 31. 32. 33
= [2. 4. 6. ... . 30. 32] [1. 3. 5. ... . 29. 31. 33]
= [(2. 1) (2. 2) (2. 3) ... (2. 15) (2. 16)] [1. 3. 5. ... . 31. 33]
= 216. (1. 2. 3. 4. ... . 15. 16). 1. 3. 5. ... . 31. 33 ...(i)
=>   33! is divisible by 216.
Further, 1. 2. 3. 4. ... . 15. 16 = (2. 4. 6...16) (1. 3. 5. ... . 15)
= (2. 1) (2. 2) (2. 3) ... (2. 8) (1. 3. 5. ... . 15)
= 28.(1. 2. 3. ... . 8) (1. 3. 5. ... . 15)
= 28 (2. 4. 6. 8) (1. 3. 5. 7) (1. 3. 5. ... . 15)
= 28. 24. (1. 2. 3. 4) (1. 3. 5. 7) (1. 3. 5. ... . 15)
= 212. (2. 4) (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15)
= 212. 2³. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15)
= 215. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15) ...(ii)
From (i) and (ii), we get
33!    = 216. 215. (1.3) (1. 3. 5. 7) (1. 3. 5. ... . 15) (1. 3. 5. ... . 33)
         = 231. (1. 3) (1. 3. 5. 7) (1. 3. 5. ... . 15) (1. 3. 5. ... . 33).
Hence, 31 is the largest value of n for which is divisible by 2 n.

Exercise

  1. Find the values of
    (i) 6!    (ii) 4! +3!    (iii) (9 -6)!
    (iv) 6! -4!    (v) 3!  . 5!
  2. Convert into factorials:
    (i) 3. 6. 9. 12. 15. 18
    (ii) 6. 7. 8. 9. 10. 11. 12
    (iii) 1. 3. 5. 7. ... (2 n -1)
    (iv) (n +1)(n +2) ... 2 n.
  3. Which of the following are true?
    (i) 4(3!) = 4!
    (ii) 3(4!) = (3 × 4)!
    (iii) 3! + 4! = (3 +4)!
    (iv) (4 -3)! = 4! - 3!
    (v) 4!. 3! = (4. 3)!
    (vi) =
  4. Find the L.C.M. and H.C.F. of 10!, 11!, 15!.
  5. Prove that 310 divides 30!. Which is highest natural number n such that 3n divides 30!
  6. >

Answers

1. (i) 720      (ii) 30      (iii) 6     (iv) 696     (v) 720
2. (i) 36. 6!    (ii) 12!/5!      (iii) 2n! / [2n. n!]       (iv) 2n! / n!
3. Only (i) is true.
4. H.C.F. is 10! and L.C.M. is 15!
5. 14