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!$$/2!$$+$$9\choose 3$$=$$220$ways
Where is the mistake? Where is the problem occuring?
 A: As Mark Saving pointed out in the comments, the error you made was not taking into account the order of the digits $1$ and $2$ in numbers in the case in which the leading digit is $1$ and the other nonzero digits are $1$ and $2$.  In that case, there are $9$ ways to place the second $1$ and eight ways to place the $2$.  With that correction, you would have obtained $220$ rather than $184$ since $P(9, 2) - \binom{9}{2} = 9 \cdot 8 - 36 = 36$.
Here is another approach to the problem.  Let $x_i$ be the $i$th digit.  Then
$$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.
