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 $\left(1+x+x^{2}+\ldots+x^{30}\right)^{2}$
$\left(x^{4}+x^{5}+x^{6}+x^{7}\right)\left(x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right)$
$x^{6}\left(\frac{1-x^{31}}{1-x}\right)^{2}\left(1+x+x^{2}+x^{3}\right)\left(1+x+x^{2}+x^{3}+x^{4}\right)$
$x^{6}\left(1-x^{31}\right)^{2}\left(1-x^{4}\right)\left(1-x^{5}\right)(1-x)^{4}$
$x^{6}\left(1-x^{4}-x^{5}+x^{9}\right)\left(1+x^{62}-2 x^{31}(1-x)^{-4}\right)$
$x^{6}\left(1-x^{4}-x^{5}+x^{9}\right)(1-x)^{-4}$
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 $=$
${ }^{24+4-1} C _{4-1}-{ }^{20+4-1} C _{4-1}-{ }^{19+4-1} C _{4-1}+{ }^{15+4-1} C _{4-1}$
$={ }^{27} C _{3}-{ }^{23} C _{3}-{ }^{22} C _{3}+{ }^{18} C _{3}=430$