Why does multinomial theorem only works for identical set of objects? I will elaborate this with an analogy, 
15 toys are to be distributed amongst 3 children , such that any child can get any number of toys, so we have to find the number of ways in which we can do so if,


*

*toys are distinct

*toys are identical
We can apply multinomial in only the latter case , why so?
 A: The answers in the two cases are clearly different, so a formula for one of the problems cannot possibly work, without change, to solve the other. A reasonable way to approach the question is to solve each of the problems, in the concrete case you mentioned. We use approaches that readily generalize.
Different toys: Call the toys $1$, $2$, and so on up to $15$ (numbers make great toys).  There are $3$ ways to decide who gets toy $1$. For each of these ways, there are $3$ ways to decide who gets toy $2$, for a total so far of $3\times 3$. For each way of making these two decisions, there are $3$ ways to decide who gets toy $3$, and so on, for a total of $3^{15}$ ways.
We could approach the problem through multinomial coefficients. The number of ways to choose $t_A$ toys to give to kid $A$, and $t_B$ toys to give to kid $B$, and $t_C$ to give to kid $C$, where $t_A+t_B+t_B=15$, is the multinomial coefficient $\binom{15}{t_A,t_B,t_C}$. To get the total number of ways, sum over all $(t_A,t_B,t_C)$. We can use the multinomial theorem to conclude that the sum is $3^{15}$. But we already had a simpler argument for $3^{15}$. 
Identical toys:  A standard approach is to count the number of ways to distribute $18$ toys among the $3$ kids, at least one toy to each kid. Then we make each kid give back a toy. Line up the $18$ toys. They determine $17$ inter-toy gaps. Put a marker into $2$ of these gaps. Give kid $A$ all the toys up to the first marker, kid $B$ the toys from the first marker to the second, and kid $C$ the rest. There are $\binom{17}{2}$ ways to choose where the markers will go, and hence $\binom{17}{2}$ ways to distribute $15$ toys among $3$ kids, where some kid(s) may get nothing. 
The argument  perhaps is not *multi*nomial,  since only the special binomial case of the multinomial is being used. But it certainly belongs to the same family of ideas.  In both cases, "nomial" ideas can be used.  
