<s>[INST] Combinatorial Problem From Book Find all the number of ways such that three elements selected from the set $\{ 1,2,3,....,4n\}$ sum of whose is divisible by $4$.
I cannot find the solution of this equation $x + y + z = 4k$. Solving this equation we can get our answer.
 [/INST] We can do the job by considering various cases. There are various ways in which the sum can be divisible by $4$. 
Case (i): The  $3$ chosen numbers are all congruent modulo $4$ (have the same remainder on division by $4$).  Then they must all be congruent to $0$ (have remainder $0$).  Since $n$ of our numbers are divisible by $4$, there are $\binom{n}{3}$ possibilities. Note that if $n\lt 3$ then $\binom{n}{3}=0$.  
Case (ii): We choose $2$ numbers congruent to $1$ modulo $4$, and $1$ congruent to $2$. That gives $\binom{n}{2}\binom{n}{1}$ ways.
Case (iii): We choose  $2$ numbers congruent to $3$, and $1$ congruent to $2$. We get the same number of ways as in (ii).
Case (iv): We choose $2$ numbers congruent to $2$, and $1$ congruent to $0$. Again, we get the same number as in (ii). 
Case (v): all the chosen numbers are incongruent modulo $4$. Then $1$ number is congruent to $1$, another to $3$, and another to $4$. That gives $\binom{n}{1}\binom{n}{1}\binom{n}{1}$ possibilities.
Finally, add up. 
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