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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
2 years, 5 months ago

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1 vote

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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!

Chinmay Sangawadekar - 2 years, 5 months ago

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

Dhruv Saxena - 2 years, 1 month ago

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You are welcome...

Chinmay Sangawadekar - 2 years, 1 month ago

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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.

Preston Kilian - 2 years, 5 months ago

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niceee....;)

Jarin Tasnim - 1 year, 11 months ago

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

Jamil Osman - 2 years, 5 months ago

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Thank you!

Yoogottam Khandelwal - 2 years, 5 months ago

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One can prove this using definition of permutations.

Shivamani Patil - 2 years, 5 months ago

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Could you expand it?

Yoogottam Khandelwal - 2 years, 5 months ago

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