Counting numbers of fruit baskets 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&#39;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.
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.
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!