The number of ways to distribute 30 identical | vedclass

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(D) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the number of candies received by children $C_{1}, C_{2}, C_{3}, C_{4}$ respectively.We have $x_{1} + x_{2} +

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$430$

Vedclass · Answer

(D) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the number of candies received by children $C_{1}, C_{2}, C_{3}, C_{4}$ respectively.We have $x_{1} + x_{2} + x_{3} + x_{4} = 30$,where $x_{1}, x_{4} \ge 0$,$4 \le x_{2} \le 7$,and $2 \le x_{3} \le 6$.Let $x_{2} = 4 + y_{2}$ where $0 \le y_{2} \le 3$,and $x_{3} = 2 + y_{3}$ where $0 \le y_{3} \le 4$.Substituting these into the equation: $x_{1} + (4 + y_{2}) + (2 + y_{3}) + x_{4} = 30 \Rightarrow x_{1} + y_{2} + y_{3} + x_{4} = 24$.The number of ways is the coefficient of $x^{24}$ in the expansion of $(1+x+x^{2}+x^{3})(1+x+x^{2}+x^{3}+x^{4})(1+x+x^{2}+\dots)^{2}$.This is the coefficient of $x^{24}$ in $(1-x^{4})(1-x^{5})(1-x)^{-4} = (1-x^{4}-x^{5}+x^{9})(1-x)^{-4}$.Using the formula for the coefficient of $x^{n}$ in $(1-x)^{-r}$ as $\binom{n+r-1}{r-1}$,we get:Coefficient $= \binom{24+4-1}{4-1} - \binom{20+4-1}{4-1} - \binom{19+4-1}{4-1} + \binom{15+4-1}{4-1}$.$= \binom{27}{3} - \binom{23}{3} - \binom{22}{3} + \binom{18}{3}$.$= 2925 - 1771 - 1540 + 816 = 430$.

The number of ways to distribute $30$ identical candies among four children $C_{1}, C_{2}, C_{3}$ and $C_{4}$ such that $C_{2}$ receives at least $4$ and at most $7$ candies,and $C_{3}$ receives at least $2$ and at most $6$ candies,is equal to

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