Suppose you have $10$ apples, $12$ bananas, and $8$ peaches, and you want to divide them into $3$ baskets containing $10$ fruit each.
In how many ways can you do this, if the fruit of each type is indistinguishable but the baskets are numbered?
(This question is a modification of A combinatorial task I just can't solve, and
I couldn't think of a simple approach to solve it other than trying to list all the possible ways of distributing the bananas in the baskets, and then the apples.
On
This is my best attempt:
There are 30 pieces of fruit. Split them into 3 baskets containing 10 pieces of fruit each
basket 1 30 ways for fruit 1 29 ways for fruit 2 ..... 21 ways for fruit 10
basket 2 20 ways for fruit 1 19 ways for fruit 2 .... 11 ways for fruit 10
basket 3 10 ways for fruit 1 9 ways for fruit 2 .... 1 way for fruit 10
Looking at it from a brute force method.
Multiply each choice together gives you the number of ways to place 30 pieces of fruit into 3 baskets of 10 each.
30*29*28.....*1 = 30!
This is not difficult using a CAS though I suspect something more clever is expected. The species here is
$$\mathfrak{S}_{=3} (\mathfrak{M}_{=10}(\mathcal{A}+\mathcal{B}+\mathcal{P})).$$
This yields
$$[A^{10} B^{12} P^8] (Z(S_{10})(A+B+P)))^3 = 1980.$$
The notation $Z(S_{10})$ refers to the cycle index of the symmetric group. We also use the fact that the cycle index corresponding to the identity acting on three slots is $Z(E_3) = a_1^3.$
Recall the recurrence by Lovasz for the cycle index $Z(S_n)$ of the multiset operator $\mathfrak{M}_{=n}$ on $n$ slots, which is $$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l}) \quad\text{where}\quad Z(S_0) = 1.$$
This makes it possible to compute these cycle indices very efficiently. We could also compute $Z(S_{10})(A+B+P)$ by iterating over partitions of the number ten into three parts, possibly zero.
Nonetheless it should be pointed out that what we have here is just algebra (no qualitative gain) however the complexity is better than iterating over all $${30\choose 10,12,8} = 3784770539550$$ assignments to the thirty slots, converting the three consecutive blocks of ten adjacent elements into sets and tracking all possibilities thus obtained. Observe that $Z(S_{10})(A+B+P)$ has just $$[z^{10}] \frac{1}{(1-z)^3} = {12\choose 2} = 66$$ terms. For the cube we get $$[z^{30}] \frac{1}{(1-z)^3} = {32\choose 2} = 496$$ terms. These values are very reasonable for a CAS.
Remark. There is no magic here, we are simply raising the generating function of the possible sets for the three slots to the third power and extracting the desired coefficient.