Step-by-step explanation:
Permutation :The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed.
[tex]nPr = \frac{n!}{(n-r)!}[/tex]
Factorial : There are n! ways of arranging n distinct objects into an ordered sequence.
Considering a situation when n = r in a permutation, nPr reduces to n!, a simple factorial of n.
Proof: 3P3 = 3!
n = 3 and r = 3
[tex]3P3 = \frac{3!}{(3-3)!}[/tex]
[tex]3P3 = \frac{3!}{(0)!}[/tex]
But 0! = 1
3P3 = 3!
Answer: Assuming we are to determine the number of distinct 3 letter groupings we can form from the letter "ABC''
without repetition of letter.
This is a permutation case and can be solved as
N = 3P3 = 3!/(3-3)!
N = 3!/0! note: 0! = 1
N = 3!
Therefore N = 3P3 = 3!
Step-by-step explanation:
permutation is the act of arranging the members of a set into a sequence or order( in permutation order is important)
Permutation can be defined as;
nPr = n!/(n-r)!
For 3P3
Assuming we are to determine the number of distinct 3 letter groupings we can form from the letter "ABC''
without repetition of letter.
This is a permutation case and can be solved as
N = 3P3 = 3!/(3-3)!
N = 3!/0! note: 0! = 1
N = 3!
Therefore N = 3P3 = 3!
