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**0! = ?**

if, 3! = 3x2x1 = 6

2! = 2x1 =2

1! = 1

0! = still 1

Stunning right!

Here are some explanations.

** Proof 1:** For **n factorial**, we can write down as

n! = n x (n-1)!

2! = 2x (2-1)! = 2x1!

1! = 1x(1-1)! = 1x0!

this proves 0! has to be 1.

**Proof 2:**

Factorial is used to compute the number of permutations. Permutations is the process of arranging the elements in a set.

3! = {1, 2, 3} { 1, 3, 2} {2, 1 , 3} {2 , 3, 1} {3, 2, 1} {3, 1 , 2}

2! = {1,2} {2,1}

1! = {1}

0! = {0}

so 0! should also have one possibility or permutation

so 0! =1

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