All 3 digit numbers the sum of whose digits is not greater than 16 My working: Let the digits of the numbers be $x,y,z$ where
\begin{align}1\leq x&\leq 9\\0\leq y,z&\leq 9\\x+y+z &\leq 16 \end{align}
I tried to solve it by making different cases here and using counting but it got really complicated and i couldnt get very far.
So how do i solve such questions?

Let $h$ denote the hundreds digit, $t$ denote the tens digit, and $u$ denote the units digit.  Since the digit sum is at most $16$, 
$$h + t + u \leq 16 \tag{1}$$
where $1 \leq h \leq 9$, $0 \leq t \leq 9$, and $0 \leq u \leq 9$.  
Let $h' = h - 1$.  Then $h'$ is a non-negative integer satisfying the inequalities $0 \leq h' \leq 8$.  Substituting $h' + 1$ for $h$ in inequality 1 yields 
\begin{align*}
h' + 1 + t + u \leq 16\\
h' + t + u \leq 15 \tag{2}
\end{align*}
Inequality 2 is an inequality in the non-negative integers. 
If we let $s = 15 - (h' + t + u)$, then we obtain the equation 
$$h' + t + u + s = 15 \tag{3}$$
which is an equation in the non-negative integers.  If we ignore, for the moment, the restrictions on $h'$, $t$, and $u$, a particular solution of equation 3 corresponds to the placement of three addition signs in a row of fifteen ones.  For instance, 
$$1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1 1 +$$
corresponds to the solution $h' = 4$, $t = 5$, $u = 6$, and $s = 0$, while 
$$1 1 1 + 1 1 1 1 + 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $h' = 3$, $t = 4$, $u = 2$, and $s = 6$.  Hence, the number of solutions of equation 3 in the non-negative integers is 
$$\binom{15 + 3}{3} = \binom{18}{3}$$
since we must select which three of the eighteen symbols (fifteen ones and three addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$.  We must exclude those solutions.   
Suppose $h' > 8$.  Then $h' \geq 9$.  Let $h'' = h' - 9$.  Then $h''$ is a non-negative integer.  Substituting $h'' + 9$ for $h'$ in equation 3 yields
\begin{align*}
h'' + 9 + t + u + s & = 15\\ 
h'' + t + u + s & = 6 \tag{4}
\end{align*}
Equation 4 is an equation in the non-negative integers with 
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions.  
Suppose $t > 9$.  Then $t \geq 10$.  Let $t' = t - 10$.  Then $t'$ is a non-negative integer.  Substituting $t' + 10$ for $t$ in equation 3 yields
\begin{align*}
h' + t' + 10 + u + s & = 15\\
h' + t' + u + s & = 5 \tag{5}
\end{align*}
Equation 5 is an equation in the non-negative integers with 
$$\binom{5 + 3}{3} = \binom{8}{3}$$
solutions.  
By symmetry, there are also $\binom{8}{3}$ solutions in which $u > 9$.
Notice that since $9 + 10 = 19 > 15$, at most one of the conditions $h' \leq 8$, $t \leq 9$, and $u \leq 9$ can be violated simultaneously in equation 3.  Hence, the number of solutions of equation 3 subject to those restrictions is 
$$\binom{18}{3} - \binom{9}{3} - 2\binom{8}{3}$$
