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 atleast 4 and atmost 7 candies, $C_3$ receives atleast 2 and atmost 6 candies, is equal to 

 $t_1+t_2+t_3+t_4=30$
Coefficient of  $x^{30}in{(1+x+x^2+\mathrm{...}{x}^{30})}^2$
 $(x^4+x^5+x^6+x^7)(x^2+x^3+x^4+x^5+x^6)$
 $x^6{\left(\frac{1-x^{31}}{1-x}\right)}^2(1+x+x^2+x^3)(1+x+x^2+x^3+x^4)$
 $x^6{(1-x^{31})}^2(1-x^4)(1-x^5){(1-x)}^{-4}$
 $x^6(1-x^4-x^5+x^9)(1+x^{62}-2x^{31}{(1-x)}^{-4})$
 $x^6(1-x^4-x^5+x^9){(1-x)}^{-4}$
Coefficient of $x^n$  in ${(1-x)}^{-t}$  is  ${}^{n+r-1}C_{r-1}$
 $\Rightarrow {}^{27}C_3-{}^{23}C_3-{}^{22}C_3+{}^{18}C_3$
 $2925-1771-1540+816=430$
  (or)
 $x_2\in [4,7],x_3\in [2,6]$
 $\Rightarrow t_1+t_2+t_3+t_4=24$
Total ways =  ${}^{24+4-1}C_{4-1}-{}^{20+4-1}C_{4-1}-{}^{19+4-1}C_{4-1}+{}^{15+4-1}C_{4-1}$
 $={}^{27}C_3-{}^{23}C_3-{}^{22}C_3+{}^{18}C_3=430$