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)
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.
Find the value of
Find n, if n! /[2! (n -2)!] and n! /[4! (n -4)!] are in the ratio 2 : 1.
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.
Prove that 33! is divisible by 216. What is the largest integer n such that 33! is divisible by 2n?
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.