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

*

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


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


*seven children get one each and the other student gets three friuts
My attempt

*

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


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


*$^8C_1 \times^{10}C_3\times7! = 4838400$
is there anything I need to check or any other ways to get theses answers ?
 A: 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 $a_2 = T_2 - \binom{2}{1} a_1 = 33$.
A: The usual convention in these problems is that people are distinguishable while fruits of the same type are not.  Under that convention, what matters is how many fruits of each type each person receives.
Also, when you consider cases, you must ensure that they are mutually exclusive and exhaustive.  Your case seven children receive one fruit each and the other child receives three fruits contains your cases seven children receive one fruit each and the other child receives three fruits.
If you approach the problem by considering how many children receive one fruit each, how many receive two fruits each, and how many receive three fruits each, you must also consider which types of fruit the children receive.  That can lead to numerous cases.
Let's distribute the mangoes first, give each child who did not receive a mango an apple, then distribute the remaining apples.  This strategy works under the assumption that fruits of the same type are indistinguishable. If each fruit were distinct, we would need a different strategy to avoid overcounting.
Six children each receive one mango:  There are $\binom{8}{6}$ ways to select the recipients of the mangoes.  Each of the remaining children must receive an apple.  That leaves two apples to distribute to eight children.  We can give both of those apples to one child in eight ways and to two different children in $\binom{8}{2}$ ways.  Hence, there are
$$\binom{8}{6}\left[\binom{8}{1} + \binom{8}{2}\right]$$
such distributions.
Exactly five children receive mangoes:  For this to occur, one of those children must receive two mangoes while each of the other four children who receives a mango receives one mango each.  There are $\binom{8}{5}$ ways to select the five children who will receive a mango and five ways to select which of those five children will receive two mangoes.  Give each of the three children who did not receive a mango an apple.  That leaves one apple to distribute to one of the eight children, which can be done in eight ways.  Hence, there are
$$\binom{8}{5}\binom{5}{1}\binom{8}{1}$$
such distributions.
Exactly four children receive mangoes:  There are two possibilities.  Either one child receives three mangoes while three other children each receive one or two children each receive two mangoes while two other children each receive one.  There are $\binom{8}{4}$ ways to select which four children will receive the mangoes.  Each of the other four children must receive an apple.  If one child will receive three mangoes, there are four ways to select that child from the four children who have been selected to receive mangoes.  If two children will each receive two mangoes, there are $\binom{4}{2}$ ways to select those children from among the four who will receive mangoes.  Hence, there are
$$\binom{8}{4}\left[\binom{4}{1} + \binom{4}{2}\right]$$
such distributions.
Total:  Since these cases are mutually exclusive and exhaustive, the number of ways six indistinguishable mangoes and four indistinguishable apples can be distributed to eight children so that each child receives at least one fruit is
$$\binom{8}{6}\left[\binom{8}{1} + \binom{8}{2}\right] + \binom{8}{5}\binom{5}{1}\binom{8}{1} + \binom{8}{4}\left[\binom{4}{1} + \binom{4}{2}\right]$$
A: You asked about correcting your counts.
The convention in problems of this type is that children are distinguishable while fruits of the same type are not.  Under this convention, what matters is how many fruits of each type each child receives.
Since there are $10$ fruits and eight children, either one child receives three fruits and each of the others each receives one fruit or two children each receive two fruits and each of the others receives one fruit.  This leads to numerous subcases, not all of which you considered.
One child receives three fruits and each of the others receives one fruit
One child receives three mangoes, three other children each receive one mango, and the remaining four children each receive one apple:  There are eight ways to select the child who receives three mangoes and $\binom{7}{3}$ ways to select which three of the other children each receive one mango.  The remaining four children must each receive an apple.  Hence, there are
$$\binom{8}{1}\binom{7}{3}$$
such distributions.
One child receives two mangoes and one apple, four other children each receive one mango, and the remaining three children each receive one apple:  There are eight ways to select the child who receives two mangoes and one apple and $\binom{7}{4}$ ways to select which four of the other children each receive one mango.  The remaining three children must each receive an apple.  Hence, there are
$$\binom{8}{1}\binom{7}{4}$$
such distributions.
One child receives one mango and two apples, five other children each receive one mango, and the remaining two children each receive one apple:  There are eight ways to select the child who receives one mango and two apples and $\binom{7}{5}$ ways to select which five of the other children each receive one mango.  The remaining two children must each receive an apple.  Hence, there are
$$\binom{8}{1}\binom{7}{5}$$
such distributions.
One child receives three apples, another child receives one apple, and each of the other six children receives one mango:  There are eight ways to select the child who receives three apples and seven ways to select the child who receives one apple.  The other six children must each receive one mango.  Hence, there are
$$\binom{8}{1}\binom{7}{1}$$
such distributions.
Two children each receive two fruits and each of the others receives one fruit
Two children each receive two mangoes, two other children each receive one mango, and the remaining four children each receive one apple:  There are $\binom{8}{2}$ ways to select the two children who each receive two mangoes and $\binom{6}{2}$ ways to select which two of the other children each receive one mango.  The remaining four children must each receive one apple.  Hence, there are
$$\binom{8}{2}\binom{6}{2}$$
such distributions.
One child receives two mangoes, another child receives one mango and one apple, three other children each receive one mango, and the remaining three children each receive an apple:  There are eight ways to select the child who receives two mangoes, seven ways to select the child who receives one mango and one apple, and $\binom{6}{3}$ ways to select which three of the other children receive one mango each.  The remaining three children must each receive an apple.  Hence, there are
$$\binom{8}{1}\binom{7}{1}\binom{6}{3}$$
such distributions.
One child receives two mangoes, another child receives two apples, four other children each receive one mango, and the remaining two children each receive one apple:  There are eight ways to select the child who receives two mangoes, seven ways to select the child who receives two apples, and $\binom{6}{4}$ ways to select which four of the other children each receive one mango.  The remaining two children must receive one apple.  Hence, there are
$$\binom{8}{1}\binom{7}{1}\binom{6}{4}$$
such distributions.
Two children each receive one mango and one apple, four other children each receive one mango, and the other two children each receive one apple:  There are $\binom{8}{2}$ ways to select the two children who will each receive one mango and one apple and $\binom{6}{4}$ ways to select which four of the other children will each receive one mango.  The remaining two children must each receive one apple.  Hence, there are
$$\binom{8}{2}\binom{6}{4}$$
such distributions.
One child receives one mango and one apple, another child receives two apples, five other children each receive one mango, and the remaining child receives one apple:  There are eight ways to select the child who receives one mango and one apple, seven ways to select the child who receives two apples, and $\binom{6}{5}$ ways to select which five of the other children each receive one mango.  The remaining child must receive the remaining apple.  Hence, there are
$$\binom{8}{1}\binom{7}{1}\binom{6}{5}$$
such distributions.
Two children each receive two apples and each of the other children receives one mango:  There are $\binom{8}{2}$ to select which two children will each receive two apples.  Each of the other children must receive one mango.  Hence, there are
$$\binom{8}{2}$$
such distributions.
Total:  Since the above cases are mutually exclusive and exhaustive, the number of admissible cases is
$$\binom{8}{1}\binom{7}{3} + \binom{8}{1}\binom{7}{4} + \binom{8}{1}\binom{7}{5} + \binom{8}{1}\binom{7}{1} + \binom{8}{2}\binom{6}{2} + \binom{8}{1}\binom{7}{1}\binom{6}{3} + \binom{8}{1}\binom{7}{1}\binom{6}{4} + \binom{8}{2}\binom{6}{4} + \binom{8}{1}\binom{7}{1}\binom{6}{5} + \binom{8}{2}$$
A: using $\color{blue}{\text{generating function method}}$,
the answer is $[x^4y^6][(\sum\limits_{i=0}^{4}\sum\limits_{j=0}^{6}x^iy^j)-1)]^8$
numberOfPeople = 8;
numberOfFruit1 = 4;
numberOfFruit2 = 6;
(Sum[x^i*y^j, {i, 0, numberOfFruit1}, {j, 0, numberOfFruit2}] - 1)^
   numberOfPeople // Expand // 
 Coefficient[#, x^numberOfFruit1*y^numberOfFruit2] &

$$
\color{red}{3948}
$$
