# 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,

1. toys are distinct

2. toys are identical

We can apply multinomial in only the latter case , why so?

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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.