In how many ways can $10$ chocolates be distributed to $3$ people such that no one gets more than $4$? $10$ chocolates are distributed to $3$ people such that no one gets more than $4$ or $A,B,C \leq 4$. How many ways can this be done?
I know how to do this if the condition is every person gets at least one. Here the condition is different. So need help.
 A: Set:


*

*$A'=4-A$

*$B'=4-B$

*$C'=4-C$


Then $A',B',C'$ must be elements of $\{0,1,2,3,4\}$ with: $$A'+B'+C'=2\tag1$$
If $A',B',C'$ are nonnegative integers satisfying this equation then automatically they will be elements of $\{0,1,2,3,4\}$ , so we can restate the problem as: find nonnegative integers $A',B',C'$ that satisfy (1). For solving this stars and bars can be applied. 
We find:  $$\binom{2+(3-1)}{3-1}=\binom42=6$$ possibilities.
A: drhab has provided you with an elegant solution to this problem based on the assumption that the chocolates are indistinguishable.  I will make the same assumption.
We wish to distribute ten chocolates to three people so that no person receives more than four chocolates.  The number of ways we can do this is equal to the number of solutions of the equation 
$$x_1 + x_2 + x_3 = 10 \tag{1}$$
in the non-negative integers subject to the restrictions that $x_1, x_2, x_3 \leq 4$.  
The number of ways ten chocolates can be distributed to three people is the number of solutions of equation 1 in the non-negative integers.  A particular solution corresponds to the placement of two addition signs in a row of ten ones.  For instance,
$$+ 1 1 1 1 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 6$, and $x_3 = 4$, while 
$$1 1 1 1 1 + 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 5$, $x_2 = 3$, and $x_4 = 2$.  Thus, the number of solutions of equation 1 in the non-negative integers is the number of ways two addition signs can be inserted into a row of ten ones, which is 
$$\binom{10 + 2}{2} = \binom{12}{2}$$
since we must choose which two of the twelve symbols (ten ones and two addition signs) will be addition signs.
However, we have counted solutions in which at least one person receives more than four chocolates.  We must exclude those cases.  
Notice that $2 \cdot 5 = 10$, so at most two of the people could receive more than four chocolates.
Suppose $x_1 > 4$.  Since $x_1$ is an integer, then $x_1 \geq 5$.  Let $y_1 = x_1 - 5$.  Then $y_1$ is a non-negative integer.  Substituting $y_1 + 5$ for $x_1$ in equation 1 yields
\begin{align*}
y_1 + 5 + x_2 + x_3 & = 10\\
y_1 + x_2 + x_3 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers with 
$$\binom{5 + 2}{2} = \binom{7}{2}$$
solutions.  There are $\binom{3}{1}$ ways to select a person to receive more than four chocolates.  Hence, the number of ways of distributing ten chocolates to three people so that one person receives more than four chocolates is 
$$\binom{3}{1}\binom{7}{2}$$ 
However, if we subtract this number from the total number of ways of distributing ten chocolates, we will have subtracted those solutions in which two people received more than four chocolates twice.  Suppose $x_1, x_2 > 4$.  Let $y_1 = x_1 - 5$; let $y_2 = x_2 - 5$.  Then $y_1, y_2$ are non-negative integers.  Substituting $y_1 + 5$ for $x_1$ and $y_2 + 5$ for $x_2$ in equation 1 yields
\begin{align*}
y_1 + 5 + y_2 + 5 + x_3 & = 10\\
y_1 + y_2 + x_3 & = 0 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with 
$$\binom{0 + 2}{2} = \binom{2}{2} = 1$$
solution (namely $y_1 = y_2 = x_3 = 0$).  Since there are $\binom{3}{2}$ ways of selecting two people to receive five chocolates each, the number of ways two people could receive more than four chocolates is 
$$\binom{3}{2}\binom{2}{2}$$
By the Inclusion-Exclusion Principle, the number of ways ten chocolates can be distributed to three people so that no person receives more than chocolates is 
$$\binom{12}{2} - \binom{3}{1}\binom{7}{2} + \binom{3}{2}\binom{2}{2} = 6$$
As a check, they are $(2, 4, 4), (4, 2, 4), (4, 4, 2), (3, 3, 4), (3, 4, 3), (4, 3, 3)$.
