The number of ways of distributing 15 apples | vedclass

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(A) Let $x_A, x_B, x_C$ be the number of apples given to persons $A, B, C$ respectively. We have $x_A + x_B + x_C = 15$ with $x_A \ge 2, x_C \ge 2$ an

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) Let $x_A, x_B, x_C$ be the number of apples given to persons $A, B, C$ respectively. We have $x_A + x_B + x_C = 15$ with $x_A \ge 2, x_C \ge 2$ and $0 \le x_B \le 5$.Let $x_A = y_A + 2$ and $x_C = y_C + 2$,where $y_A, y_C \ge 0$.Substituting these into the equation: $(y_A + 2) + x_B + (y_C + 2) = 15 \implies y_A + x_B + y_C = 11$.The number of ways is the coefficient of $x^{11}$ in the expansion of $(1+x+x^2+\dots)^2(1+x+x^2+x^3+x^4+x^5)$.This is the coefficient of $x^{11}$ in $(1-x)^{-2} 	imes \frac{1-x^6}{1-x} = (1-x^6)(1-x)^{-3}$.Expanding $(1-x^6) \sum_{n=0}^{\infty} \binom{n+3-1}{3-1} x^n = (1-x^6) \sum_{n=0}^{\infty} \binom{n+2}{2} x^n$.The coefficient of $x^{11}$ is $\binom{11+2}{2} - \binom{(11-6)+2}{2} = \binom{13}{2} - \binom{7}{2}$.$= \frac{13 	imes 12}{2} - \frac{7 	imes 6}{2} = 78 - 21 = 57$.

The number of ways of distributing $15$ apples to three persons $A, B, C$ such that $A$ and $C$ each get at least $2$ apples and $B$ gets at most $5$ apples is

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