How many integer solutions are there to the equation : $$ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 29 $$
$$ x_{1}\geq1, x_{2}\geq2, x_{3}\geq3, 10 \geq x_{4}\geq0, x_{5} > 5, x_{6}\geq -2$$
Could anyone explain me how to approach that ? Specially when it comes to the restriction of  $x_{4}$.

Substitute $y_1 = x_1 -1$, $y_2 = x_2 -2$, $y_3 = x_3-3$, $y_4 = x_4$, $y_5 = x_5 - 5$ and $y_6 = x_6 + 2$ to get
$\;\;\;
    y_1 + y_2 + y_3 + y_4 + y_5 + y_6  = 20$ with $y_i\ge0$ for $1\le i\le6$ and $y_4\le10$.
We have to count how many ways to place 20 identical balls in 6 boxes, with the restriction that there are at most 10 balls in the 4th box.
Without any restrictions, there are $\binom{25}{5}$ ways to place the balls in the boxes, since this is the number of ways to arrange 20 balls and 5 dividers; and we have to subtract the number of distributions with 11 balls in box 4, which is equal to $\binom{14}{5}$ since there are 9 balls remaining and 5 dividers. 
Therefore the number of solutions is given by
$\binom{25}{5}-\binom{14}{5}$.
