no. of onto functions from A to A Let 

$$A =\{1,2,3,4,5,6,7\}\quad and \quad f:\mathbb{A}\rightarrow \mathbb{A}.$$ 

What is the no. of onto functions in which at least $3$ elements of $A$ have self Images?
I did not understand the Language of Question

Here is a very tedious approach:
Let $\eta_n$ be the number of bijections of $\{1,...,n\}$ that have no fixed point (that is, $f(x) \neq x$ for all $x$). We note that $\eta_1 = 0$, and by inspection $\eta_2 = 1, \eta_3 = 2$. Computing $\eta_4$ is a little more work, enumerating gives $\eta_4 = 9$ (the possibilities are $(2,1,4,3), (2,3,4,1)$, $(2,4,1,3)$, $(3,1,4,2)$, $(3,4,1,2)$, $(3,4,2,1)$, $(4,1,2,3)$, $(4,3,1,2)$, $(4,3,2,1)$).
Consider the number of bijections that have exactly $k$ fixed points. We need to consider $k=3,...,7$.
Note that there are 6 fixed points iff there are 7 fixed points, and there is exactly one of these. So we just need to consider $k=3,4,5$ and add one.
First choose $k \in \{3,4,5\}$ points, there are $\binom{7}{k}$ of these. We have $\eta_{7-k}$ possibilities for the remaining $7-n$ numbers. Hence there are  $\binom{7}{k} \eta_{7-k}$ of these.
So the number we want is $1+\sum_{k=3}^5 \binom{7}{k} \eta_{7-k} = 1 + \binom{7}{3}9 + \binom{7}{4}2 + \binom{7}{5}1 = 407$.
