Answer:
Given
S = {1, 2, 3, 4, ?.. n}
Let
f : S S be an onto
function
Since
Rf = S, therefore f is also one-one. Thus f is both one-one onto
function. Therefore number of functions is equal to number of arrangement of n
numbers taken all at a time i.e.,
So,
there are
onto
functions from set {1,2, 3, 4, ?., n} to itself.
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