You have to distribute each type of fruit separately.
Let $a_k$, $1 \leq k \leq 3$, be the number of apples distributed to the $k$th person. Then
$$a_1 + a_2 + a_3 = 5 \tag{1}$$
The number of ways the apples can be distributed is the number of solutions of equation 1 in the nonnegative integers. A particular solution corresponds to the placement of two addition signs in a row of five ones. For instance,
$$1 1 + 1 + 1 1$$
corresponds to the solution $a_1 = 2$, $a_2 = 1$, and $a_3 = 2$, while
$$+ 1 1 1 + 1 1$$
corresponds to the solution $a_1 = 0$, $a_2 = 3$, and $a_3 = 2$. Therefore, the number of such solutions is the number of ways two addition signs can be inserted into a row of five ones, which is
$$\binom{5 + 2}{2} = \binom{7}{2}$$
since we must choose which two of the seven symbols (five ones and two addition signs) will be addition signs.
The number of ways the mangoes can be distributed is the number of solutions in the nonnegative integers of the equation
$$m_1 + m_2 + m_3 = 4 \tag{2}$$
where $m_k$, $1 \leq k \leq 3$, is the number of mangoes distributed to the $k$th person. The number of such solutions is
$$\binom{4 + 2}{2} = \binom{6}{2}$$
The number of ways the oranges can be distributed is the number of solutions in the nonnegative integers of the equation
$$o_1 + o_2 + o_3 = 3 \tag{3}$$
where $o_k$, $1 \leq k \leq 3$, is the number of oranges distributed to the $k$th person. The number of such solutions is
$$\binom{3 + 2}{2} = \binom{5}{2}$$
To find the number of ways of distributing the three types of fruit, multiply the above results.
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct boxes?
You correctly found that $4$ indistinguishable oranges can be placed in $5$ distinct boxes in
$$\binom{4 + 5 - 1}{5 - 1} = \binom{8}{4}$$
ways. Since there are five ways each of the six apples can be placed in a box, there are $5^6$ ways to distribute the apples. Hence, the number of possible distributions is
$$5^6\binom{8}{4}$$
As for the other formula you considered, if you wanted to arrange four indistinguishable dividers and six different apples in a row, you could choose four of the ten positions for the dividers in $\binom{10}{4}$ ways, then arrange the six apples in the six remaining positions in $6!$ ways, which yields
$$\binom{10}{4}6! = \frac{10!}{4!6!} \cdot 6! = \frac{10!}{4!} = \frac{(6 + 5 - 1)!}{(5 - 1)!}$$
However, we do not care about the order of the apples within each box, so this formula yields too large a number for the number of ways of distributing the apples.
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct so that there are two pieces of fruit in each box?
The restriction that exactly two pieces of fruit are placed in each box means that we have three cases to consider:
- Two oranges apiece are placed in each of two boxes.
- Two oranges are placed in one box and one orange apiece is placed in each of two other boxes.
- Each orange is placed in a distinct box.
Two oranges apiece are placed in each of two boxes: Choose which two of the five boxes each receive two oranges. Choose which two of the six apples are placed in the leftmost empty box. Choose which two of the remaining four apples are placed in the leftmost remaining empty box. Place the final two apples in the remaining empty box.
There are $$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ such distributions.
Two oranges are placed in one box and one orange apiece is placed in each of two other boxes: Choose which of the five boxes receives two oranges. Choose which two of the remaining four boxes each receive one of the other two oranges. Choose which of the six apples is placed in the leftmost box with one apple and which of the remaining five apples is placed in the other box with one apple. Choose which two of the remaining four apples is placed in the leftmost empty box. The other two apples must be placed in the remaining empty box.
There are $$\binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$ such distributions.
Each orange is placed in a distinct box: There are $\binom{5}{4}$ ways to select which four of the boxes each receives one orange. There are $\binom{6}{2}$ ways to select which two apples are placed in the empty box. There are $4!$ ways to distribute the remaining four apples to the four boxes that each contain one orange.
There are $$\binom{5}{4}\binom{6}{2}4!$$ such distributions.
Since these cases are mutually exclusive and exhaustive, the number of ways to distribute four indistinguishable oranges and six distinct apples in such a way that two pieces of fruit are placed in each box is found by adding the results for each case.
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{6}{2}4!$$
Best Answer
You can give any combo of $0-5$ apples, $0-10$ mangoes to person $A$
Thus $6\times11 = 66$ ways (the balance needed to make $15$ will be oranges)