How many ways can we distribute 10 toys, 5 balls and 7 rings to 3 children?

Question

Well, this question has different solution depending on whether the toys, balls and rings are identical among themselves or not. My simple query is that by default whether we should consider the toys, balls and rings as identical ones or distinct among themselves?

Answer 1

By default, in the beginning, everything is labelled (distinct) and moreover, ordered. Only later one may talk about types, that are classes of equivalence.

For example, if I have to choose one red ball from three red balls, the balls are labeled and the problem has the solution $3$.

For the unlabeled case we have only one way to choose among {x, x, x} which is {x} and the solution is $1$. Above, if the labeling status is not specified, there are sixteen problems in one.

I will approach one of the sixteen cases : let kids and toys be unlabeled and ball and rings labelled.

i) we partition the labeled items into 1, 2, or three parts. I do this by mean of e.g.f.

a) for all items to one kid, we are interested in the coefficient $(5,7)$ of $(exp(x+y)-1)$ which is $1$.

b) for all items distributed to two (un-ordered, unlabeled) kids , the egf is $(exp(x+y)-1)^2 \over2!$ that gives $2047$ when considering the coefficient $(5,7)$

c) for three parts, $(exp(x+y)-1)^3 \over 3!$ gives $86526$.

ii)

in the case c) the kids are, for each repartition, uniquely identified by their collection of items. Thus we are interested in ordered partitions of number $10$ in $3$ parts which is $66$.

In the case b) yet again two kids are determined and the third is also, and we get another $66$ cases for each repartition of the labeled items

in the case a) with one determined kid and two identical, we have to fill a structure $x-\{x,x\}$ with ten non-distinct items.

$10 - \{0,0\}$

$9-\{1,0\}$

$8-\{2,0\} \ \ \ 8-\{1,1\}$

$7-\{3,0\} \ \ \ 7-\{2,1\}$

...

the total is $1+1+2+2+3+3+4+4+5+5+6 = 36$

$1 \times 36 + 2047\times 66 + 86526 \times 66 = 5,845,854 $