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Best answer
29 votes
29 votes
There are $n!$ one to one function possible from a set of $n$ elements to itself.

i.e., $P\binom{n}{n} = n!$
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20 votes
20 votes

f : A---> A

∣A∣ = n

The first element of the domain has n choices for mapping,2nd element has (n-1) ,3rd element has (n-2) choices and so on.

So, total number of one-to-one functions = n ⨉(n-1)⨉(n-2)⨉(n-3).........⨉1 = n!

The correct answer is n!

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