The probability that each child gets at least one chocolate is? 
5 different chocolates are to be distributed among 4 children.The
  probability that each child gets at least one chocolate is ?

Total number of ways is $4^5$.Got that.After that what to do?
 A: We count the number of "favourables." The numbers involved are very small, so we use a counting procedure that would be inefficient for larger numbers.
The lucky child can be chosen in $\binom{4}{1}$ ways. For each such way, the chocolates she gets can be chosen in $\binom{5}{2}$ ways. And for every way of doing this, there are $3!$ ways to distribute the rest of the chocolates, one to each remaining child.
Remark: The total number of ways to distribute the chocolates is $4^5$, not $5^4$.  For each of chocolates $C_1$ to $C_5$, there are $4$ ways to choose the child it will be given to.
Added: We sketch the standard Inclusion/Exclusion approach to the problem. There are $4^5$ ways to distribute the chocolates with no restriction. We now count the bad ways, in which at least one sad little child gets nothing. Call the children A, B, C, D. There are $3^5$ ways to distribute the chocolates so that A gets nothing, for we are distributing $5$ chocolates among $3$ kids. The same is true for B getting nothing, and so on. So our first estimate of the number of bad distributions is $4\cdot 3^5$, which we write as $\binom{4}{1}3^5$. 
However, we have double-counted the number of ways in which, for example, A and B both get nothing. There are $\binom{4}{2}$ ways to choose two children to get nothing, and for each way there are $2^5$ ways to distribute the chocolates among the other two kids.
So  our second estimate for the number of bad distributions is $\binom{4}{1}3^5-\binom{4}{2}2^5$. However, we have taken away too much, for we have removed once too many times the $\binom{4}{3}1^5$ ways in which all the chocolates go to one kid. We conclude that the number of bad distributions is
$$\binom{4}{1}3^5-\binom{4}{2}2^5+\binom{4}{3}1^5.$$
Subtract this from the total $4^5$ to find the number of good distributions.
The above idea works in general. 
A: We can say that 5/4=1
1+1(the given number in the question-at least one-)=2
so we say that there would be a child who will get two chocolate which means others have one chocolate.
A: Suppose you have $n$ chocolates and $m$ children.
Let $N_j$ be the set of arrangements that miss child $j$.
There are $\binom{m}{k}$ intersections of $k$ of the $N_j$. There are $(m-k)^n$ elements in each intersection of $k$ of the $N_j$. Thus, using Inclusion-Exclusion the number of arrangements that miss some child is
$$
\begin{align}
\left|\bigcup_jN_j\right|
&=\sum_j\left|N_j\right|-\sum_{j_1\lt j_2}\left|N_{j_1}\cap N_{j_2}\right|+\sum_{j_1\lt j_2\lt j_3}\left|N_{j_1}\cap N_{j_2}\cap N_{j_3}\right|-\dots\\
&=\sum_{k=1}^m(-1)^{k-1}\binom{m}{k}(m-k)^n
\end{align}
$$
Therefore, the probability that each child gets a candy is
$$
\frac1{m^n}\sum_{k=0}^m(-1)^k\binom{m}{k}(m-k)^n
$$
In the case given, $n=5$ and $m=4$:
$$
\frac1{4^5}\sum_{k=0}^4(-1)^k\binom{4}{k}(4-k)^5=\frac{15}{64}
$$
