How many ten digit numbers have the sum of their digits equal to $4$? How many ten digit numbers have the sum of their digits equal to $4$?
Well, my solution goes like this:

The first digit of the $10$ digit number cannot be $0$.
It can be $1,2,3$ or $4$ only .
If the first digit is $1$ then we can have $3$ other digits as $1$ so as the sum of digit is $4$ . This can be done in $9\choose3$ ways . Now when the 1st digit is $1$ then we can have two digits $2$ and $1$ as another case such as sum of digit remains $4$. So, this can be done in $9\choose 2$ ways . Now, another case can be the one when 1st digit is $4$ but another digit is $3$ . This can be done in $9\choose 1$ ways . So the number of ways when 1st digit is $1$ is $9\choose3$$+$$9\choose2$$+$$9\choose1$ways.
Now, if the 1st digit is $2$ then the other two digits can be $1$ each . This can done in $9\choose2$ways . If the 1st digit is $2$ the other digit can be $2$ . This can be done in $9\choose1$ways.The total of ways this can be done is $9\choose2$$+$$9\choose1$ ways. If the 1st digit is $3$ then another digit among those $9$ digits must be $1$.this can be done in $9\choose1$ways. If the 1st digit is $4$ then all the other digits are zero . This can be done in $1$ way only. So, the total number of ways in which the sum of digits can be $4$ in a $10$ - digit number is $9\choose3$$+$$9\choose2$$+$$9\choose1$$+$$9\choose2$$+$$9\choose1$$+$ $9\choose1$$+$$1$ ways$ =184$ways

However the answer in the book is given as: $1+2$$9\choose 1$$+$$9\choose1$$+$$9\choose2$$3!$$/2!$$+$$9\choose 3$$=$$220$ways
Where is the mistake? Where is the problem occuring?

As Mark Saving pointed out in the comments, the error you made was not taking into account the order of the digits $1$ and $2$ in numbers in the case in which the leading digit is $1$ and the other nonzero digits are $1$ and $2$.  In that case, there are $9$ ways to place the second $1$ and eight ways to place the $2$.  With that correction, you would have obtained $220$ rather than $184$ since $P(9, 2) - \binom{9}{2} = 9 \cdot 8 - 36 = 36$.
Here is another approach to the problem.  Let $x_i$ be the $i$th digit.  Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} = 4 \tag{1}$$
Since the leading digit cannot be zero, the integer $x_1 \geq 1$.  Each of the remaining variables represents a nonnegative integer.
Let $x_1' = x_1 - 1$. Then $x_1'$ is also a nonnegative integer.  Substituting $x_1' + 1$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 4\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers.  A particular solution of equation 2 corresponds to the placement of $10 - 1 = 9$ addition signs in a row of $3$ ones.  For instance,
$$+ + 1 + + + + 1 1 + + +$$
corresponds to the solution $x_1' = x_2 = 0$, $x_3 = 1$, $x_4 = x_5 = x_6 = 0$, $x_7 = 2$, $x_8 = x_9 = x_{10} = 0$ (in which case, the original number was $10,010,002,000$ since $x_1 = x_1' + 1$).  The number of solutions of equation 2 is the number of ways we can insert $10 - 1 = 9$ addition signs in a row of $3$ ones, which is
$$\binom{3 + 10 - 1}{10 - 1} = \binom{12}{9} = 220$$
since we must choose which nine of the twelve positions required for three ones and nine addition signs will be filled with addition signs.
