Counting using permutation and combination How many solutions are there to the equation
 $x_1+x_2+x_3+x_4+x_5=21$,
where $x_i,i=1,2,3,4,5$, is a nonnegative integer such
that
$  0 ≤ x_1 ≤ 3$, $1 ≤ x_2  < 4$, and $x_3 ≥ 15$?
I tried it .My Approach-:
$ x_3=x_3'+15 \implies x_1+x_2+x_3'+15+x_4+x_5=21 \implies x_1+x_2+x_3'+x_4+x_5=6 \implies C(5+6-1,6)$
but stuck at finding 
$0 ≤ x_1 ≤3$
$1 ≤ x_2 ≤4$
please help!!!

Let $y_3 = x_3 -15$. We need the number of solutions to
    $x_1+x_2+y_3+x_4+x_5 = 21 - 15$ with $y_3 \geq 0$, $0 \leq x_1 \leq 3$ and $1 \leq x_2 <4$. The number of solutions is the coefficient of $x^6$ in 
    \begin{align*}
 (1+x+x^2+x^3)&(x+x^2+x^3)(1+x+x^2+\cdots)(1+x+x^2+\cdots)(1+x+x^2+\cdots)\\
 &= x(1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)^3 \\
 &= x\left(\frac{1-x^4}{1-x}\right)(1+x+x^2)\left(\frac{1}{1-x}\right)^3\\
 &= x(1-x^4)(1+x+x^2)(1-x)^{-4} \\
 &=x(1-x^4)(1+x+x^2)\left(1+4x+\binom{5}{2}x^2+\binom{6}{3}x^3+ \cdots\right)\\
 &= x(1+x+x^2-x^4-x^5-x^6)\left(1+4x+\binom{5}{2}x^2+\binom{6}{3}x^3+ \cdots\right)\\
 \end{align*}
    Hence the coefficient is 
    \begin{align*}
 -1-4+\binom{6}{3}+\binom{7}{4}+\binom{8}{5} = 106
 \end{align*}
