The number of ways to distribute 30 identical candies among four children  $C_1,C_2,C_3$  and  $C_4$  so that  $C_2$  receives at least 4 and at most 7 candies, $C_3$  receives at least 2 and at most 6 candies, is equal to¬¬¬--------

Let child  $C_i$  receive  $x_i$  number of candies.
$x_1+x_2+x_3+x_4=30;$  where  $x_1,x_4\ge 0$  and  $x_2\in [4,7]$  and  $x_3\in [2,6]$
Number of integral solutions 
= Coefficient of  $x^{30}in {(1+x+x^2+\mathrm{...}+x^{24})}^2$ 
 $(x^4+x^5+x^6+x^7)(x^2+x^3+x^4+x^5+x^6)$
= Coefficient of $x^{24}$   in  ${(1+x+x^2+\mathrm{....}+x^{24})}^2$
 $(1+x+x^2+x^3)(1+x+x^2+x^3+x^4)$
= Coefficient of  $x^{24}$  in  ${\left(\frac{1-x^{25}}{1-x}\right)}^2\left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^5}{1-x}\right)$
= Coefficient of  $x^{24}$   in  $(1-2x^{25}+x^{50})(1-x^4-x^5+x^9){(1-x)}^{-4}$
= Coefficient of  $x^{24}$  in  $(1-x^4-x^5+x^9)(1+{}^{4}C_{1}x+{}^{5}C_2{x}^2+{}^{6}C_3{x}^3+\mathrm{.....})$
 $={}^{27}C_3-{}^{23}C_3-{}^{22}C_3+{}^{18}C_3$
= $2925-1771-1540+816=430$