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
 A: You really do want to use an inclusion-exclusion argument. First notice that a function from $A$ onto $A$ is just a permutation of $A$/
Suppose that $F$ is a $3$-element subset of $A$, and that $f$ is a function from $A$ onto $A$ leaving each member of $F$ fixed; $f$ can permute $A\setminus F$ arbitrarily, and since $|A\setminus F|=4$, there are $4!$ permuations of $A\setminus F$. Thus, there are $4!$ permutations of $A$ that leave the members of $F$ fixed. There are $\binom73$ $3$-element subsets of $A$, each of which is left pointwise fixed by $4!$ permutations of $A$, so to a first approximation there are $\binom734!$ permuations of $A$ that fix at least $3$ members of $A$.
Unfortunately, this calculation overcounts. Consider, for instance, the identity function $\mbox{id}_A$ on $A$ that fixes every member of $A$: $\mbox{id}_A$ fixes each $3$-element subset of $A$, so we’ll count it once for each of the $\binom73$ $3$-element subsets of $A$. In other words, we’ll count it $\binom73=35$ times! Even a permutation of $A$ that fixes only $4$ points gets counted $\binom43=4$ times, once for each $3$-element subset of its set of fixed points. The inclusion-exclusion principle is a way to compensate for this overcounting.
As I said in the previous paragraph, every permutation of $A$ that fixes $4$ points gets counted $4$ times in our first approximation; we want to count it only once, so we need to subtract it $3$ times. $A$ has $\binom74$ $4$-element subsets, and there are $(7-4)!=3!$ permutations fixing (at least) the $4$ points of any one of those subsets, so to a first approximation there are $\binom743!$ permutations of $A$ fixing (at least) $4$ points, and our second approximation  is $$\binom734!-3\binom743!\;.$$
Now consider a permutation of $A$ with $5$ fixed points: it was counted $\binom53=10$ times in the first term and $3\cdot\binom54=5$ times in the second term, so it’s been counted a net total of $10-3\cdot5=-5$ times. We want to count it only once, so we should add it $6$ times. We can use the same argument that we’ve already used twice to find that to a first approximation there are $\binom752!$ permutations of $A$ with (at least) $5$ fixed points and improve our approximation to the answer to the original problem to
$$\binom734!-3\binom743!+6\binom752!\;.\tag{1}$$
Finally, any permutation of $A$ that fixes $6$ points necessarily fixes all $7$ and is $\mbox{id}_A$. This permutation was counted $\binom73$ times as fixing $3$ points, $\binom74$ times as fixing $4$ points, and $\binom75$ times as fixing $5$ points, so it’s counted 
$$\binom73-3\binom74+6\binom75=56$$
times in $(1)$. We want it counted only once, so our final result is
$$\binom734!-3\binom743!+6\binom752!-55=407$$
permutations of $A$ with at least $3$ fixed points.
(All of this is assuming that I’ve made no silly errors.)
A: 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 $1+\sum_{k=3}^5 \binom{7}{k} \eta_{7-k} = 1 + \binom{7}{3}9 + \binom{7}{4}2 + \binom{7}{5}1 = 407$.
A: Another approach would be to take all onto functions from A to A, and then
subtract the ones which fix less than 3 elements:
Since there are $\binom{7}{i}D_{7-i}$ functions which fix exactly i elements for $0\le i\le2$, this gives
$7!-\binom{7}{2}D_{5}-\binom{7}{1}D_{6}-D_{7}=5040-21(44)-7(265)-1854=407$
(where $D_{j}$ is the number of derangements of a set with $j$ elements).
