Out of 5 apples,10 mangoes,and 15 oranges,15 | vedclass

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(A) Let the two persons be $P_1$ and $P_2$. We need to distribute $15$ fruits such that the number of apples $a \le 5$,mangoes $m \le 10$,and oranges

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(A) Let the two persons be $P_1$ and $P_2$. We need to distribute $15$ fruits such that the number of apples $a \le 5$,mangoes $m \le 10$,and oranges $o \le 15$.The generating function for one person is $f(x) = (1+x+x^2+x^3+x^4+x^5)(1+x+x^2+...+x^{10})(1+x+x^2+...+x^{15})$.Since there are two persons,the total number of ways is the coefficient of $x^{15}$ in $[f(x)]^2$.However,the problem implies distributing $15$ fruits total between two people. This is equivalent to finding the coefficient of $x^{15}$ in the expansion of $(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+...+x^{10})(1+x+x^2+...+x^{15})$.Using the formula for geometric series: $f(x) = \frac{1-x^6}{1-x} \cdot \frac{1-x^{11}}{1-x} \cdot \frac{1-x^{16}}{1-x} = (1-x^6)(1-x^{11})(1-x^{16})(1-x)^{-3}$.We need the coefficient of $x^{15}$ in $(1 - x^6 - x^{11} + x^{17} + ...)(1 + 3x + 6x^2 + ... + \binom{n+3-1}{3-1}x^n + ...)$.Coefficient of $x^{15} = 1 \cdot \binom{15+3-1}{3-1} - 1 \cdot \binom{15-6+3-1}{3-1} - 1 \cdot \binom{15-11+3-1}{3-1}$.$= \binom{17}{2} - \binom{11}{2} - \binom{6}{2} = 136 - 55 - 15 = 66$.

Out of $5$ apples,$10$ mangoes,and $15$ oranges,$15$ fruits are to be distributed between two persons. Find the total number of ways of distribution.

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