Question

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

• A. 205
• B. 615
• C. 510
• D. 430

Answer 1

Answer: (D)

$t_1+t_2+t_3+t_4=30$
Coefficient of $x^{30}$ in ${\left(1+x+x^2+\dots +x^{30}\right)}^2$
$\left(x^4+x^5+x^6+x^7\right)\left(x^2+x^3+x^4+x^5+x^6\right)$
$x^6{\left(\frac{1-x^{31}}{1-x}\right)}^2\left(1+x+x^2+x^3\right)\left(1+x+x^2+x^3+x^4\right)$
$x^6{\left(1-x^{31}\right)}^2\left(1-x^4\right)\left(1-x^5\right)(1-x{)}^4$
$x^6\left(1-x^4-x^5+x^9\right)\left(1+x^{62}-2x^{31}(1-x{)}^{-4}\right)$
$x^6\left(1-x^4-x^5+x^9\right)(1-x{)}^{-4}$
Coefficient of $x^n$ in $(1-x{)}^{-r}$ 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 $=$
${}^{24+4-1}{C}_{4-1}-{}^{20+4-1}{C}_{4-1}-{}^{19+4-1}{C}_{4-1}+{}^{15+4-1}{C}_{4-1}$
$={}^{27}{C}_3-{}^{23}{C}_3-{}^{22}{C}_3+{}^{18}{C}_3=430$