# 0!=1?

Let n be any integer

so,$$n!$$=$$1\times2\times3\times\dots\times (n-1)\times n$$

and, $$n-1!$$=$$1\times2\times3\times\dots\times (n-1)$$

now, multiplying and dividing R.H.S. of $$n-1!$$ by n, we get

$\frac { 1\times 2\times 3\times ....\times n-1\times n }{ n }$

which is actually $\frac { n! }{ n }$

now, put $$n=1$$ in $$n-1!$$, we get

$$1-1!$$=$$0!$$ on L.H.S. and $\frac { 1! }{ 1 }$=$$1$$ in R.H.S.

so, $$0!$$=$$\boxed1$$

Please give some more solutions to this..Thank you!

Note by Yoogottam Khandelwal
3 years, 5 months ago

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If n! is defined as the product of all positive integers from 1 to n, then: 1! = 1*1 = 1

2! = 1*2 = 2

3! = 123 = 6

4! = 123*4 = 24 ...

n! = 123...(n-2)(n-1)n

and so on.

Logically, n! can also be expressed n*(n-1)! .

Therefore, at n=1, using n! = $$n*(n-1)!$$

1! = 1*0! which simplifies to 1 = 0!

- 3 years, 4 months ago

The most comprehensible explanation by far...! Thank you :)

- 3 years ago

You are welcome...

- 3 years ago

Using words alone you can determine 0!=1. factorials are used to tell us how many ways there are to arrange n amount of objects. if there are 0 objects there is only 1 way to arrange it. The proof is a nice way to show it but for basic understanding of what a factorial is and does words can help people who are learning it for the first time.

- 3 years, 4 months ago

niceee....;)

- 2 years, 10 months ago

Very good explanation of 0!=1 by Yoogottam Khandelwal.

- 3 years, 4 months ago

Thank you!

- 3 years, 4 months ago

One can prove this using definition of permutations.

- 3 years, 4 months ago

Could you expand it?

- 3 years, 4 months ago