Question

10. Create a scenario to explain why 3P3=3!.

Answer 1

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!

Answer 2

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.

                                  $nPr = \frac{n!}{(n-r)!}$

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

                                   $3P3 = \frac{3!}{(3-3)!}$

                                   $3P3 = \frac{3!}{(0)!}$

But 0! = 1

                                    3P3 = 3!