Let n be any integer
so,n!=1×2×3×⋯×(n−1)×n
and, n−1!=1×2×3×⋯×(n−1)
now, multiplying and dividing R.H.S. of n−1! by n, we get
n1×2×3×....×n−1×n
which is actually nn!
now, put n=1 in n−1!, we get
1−1!=0! on L.H.S. and 11!=1 in R.H.S.
so, 0!=1
Please give some more solutions to this..Thank you!
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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!
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The most comprehensible explanation by far...! Thank you :)
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You are welcome...
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.
One can prove this using definition of permutations.
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Could you expand it?
Very good explanation of 0!=1 by Yoogottam Khandelwal.
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Thank you!
niceee....;)