Number of ways of distributing 4 apples and 6 mangoes to 8 children so that each child receives at least one fruit

combinatoricsgenerating-functionspermutations

Question

Six mangoes and four apples are to be distributed among eight children so that each child receives at least one fruit. Find the number of different ways in which

  1. six children get one each and out of the remaining two children one gets two mangoes and the other gets two apples.

  2. seven children get one each and the other student gets three mangoes

  3. seven children get one each and the other student gets three friuts

My attempt

  1. $^8C_1 \times^6C_2\times^7C_1\times^4C_2\times6! =3628800 $

  2. $^8C_1 \times^6C_3\times7! = 806400$

  3. $^8C_1 \times^{10}C_3\times7! = 4838400$

is there anything I need to check or any other ways to get theses answers ?

Best Answer

It is natural to assume that children are distinguishable while fruits of the same kind are not. Then the answer is the number of all non-negative integer solutions of the system:

$$ \left\{\begin{matrix} x_1 + x_2 +...+x_8= 4\\ y_1 + y_2+...+ y_8 = 6\\ 1\leq x_i + y_i \quad i=1,...,8\\ \end{matrix}\right.$$

Lets $a_8$ be the answer. To obtain this $a_8$, first, we neglect the inequality constraints, so the total number of solutions would be $T_8 = \binom{11}{7} \times \binom{13}{7}$. Then, $a_8$ is equal to $T_8$ subtracted by the number of solutions where exactly $k$ inequality constraints fail $( x_i + y_i = 0 )$ for all possible various $k$. Obviously, $k$ is running from $1$ to $7$. This means

$$ a_8 = T_8 - \sum_{k=1}^{7} \binom{8}{k} a_{8-k} $$

Now applying the same argument on all $a_7, a_6 ,,, a_2$ we can write $a_8$ only in terms of $a_1 = 1$. For example $a_2 = T_2 - \binom{2}{1} a_1 = 33$.

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