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One mapping is selected at random from all the mappings of the set `A={1,2, 3.. n}` into itself. The probability that eh mapping selected is one to one is a. `1/(n^n)` b. `1/(n !)` c. `((n-1)!)/(n^(n-1))` d. none of these

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The mapping is from set A to A. So, domain and codomain of function are A.
Now, each pre-image in set A can be assigned any one of the images from set A.
So, total number mappings = `n xx n xx n xx … xx n` (n times) = `n^(n)`
Number of mappings which are one-one
`= nxx (n-1) xx (n-2)xx ... xx 2 xx1 = n!`
So, required probability = `(n!)/(n^(n)) = ((n-1)!)/(n^(n-1))`
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