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How many one-to-one functions are there from a set $A$ with $n$ elements onto itself?

There are n! one to one function possible with n elements to itself.. I.e. $P\binom{n}{n}$ = n!
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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! .

answered ago by Active (1.7k points)