In How Many Ways Can 7 Toys Be Given To 3 Children When Each Child is eligible for all the toys

Question

In How Many Ways Can 7 Toys Be Given To 3 Children When Each Child is eligible for all the toys?

Answer is 2187 but How???

Answer 1

There is a natural bijection between each way to give the toys to the children and the set of functions with domain $\{\text{toy1, toy2, toy3, ... toy7}\}$ and codomain $\{\text{child1, child2, child3}\}$.

The set of functions with domain $\{\text{toy1, toy2, toy3, ... toy7}\}$ and codomain $\{\text{child1, child2, child3}\}$ is represented as $\{\text{child1, child2, child3}\}^{\{\text{toy1, toy2, toy3, ... toy7}\}}$ and has cardinality $|\{\text{child1, child2, child3}\}|^{|\{\text{toy1, toy2, toy3, ... toy7}\}|}=3^7=2187$

Alternatively worded, we approach via multiplication principle. Pick which child receives toy1, pick which child receives toy2, etc... There are three options for each toy. Multiplying the number of options available for each step, we have a count of $3\times 3\times\dots\times 3 = 3^7=2187$

Answer 2

Each toy may go to any child with same probability. Hence the number of possibilities for each toy is $3$. This is true for each toy, and there are $7$ toys, hence total number of ways$=3^7=2187$.