2929506

How to solve 4 apples, 5 oranges and 6 bananas in 4 baskets?
how-to-solve-4-apples-5-oranges-and-6-bananas-in-4-baskets

( I have found a recurrence formula here : https://www.cse.iitk.ac.in/users/rgurjar/projects/CycleIndexPolynomial.pdf)

The species formula for one non-empty basket is

$$oneBasket = Ens_0^4 \ (A) \cdot Ens_0^5 \ (O) \cdot Ens_0^6 \ (B) -1 $$

where $$E_0^k$$ are sets with at most $$k$$ elements. A set of four non-empty baskets is: $$Ens_4 \ (oneBasket)$$

Since the parts of a partition must be non-empty, we will have to calculate four cases, from one non-empty basket to four non-empty baskets.

After writing the cycle index we will extract some coefficients of $$ \ {a^4\over 4!}{o^5\over 5!}{b^6\over 6!} \ $$ for the e.g.f. and $$ \ a^4o^5b^6 \ $$ for the GF, the types generating function.

1) We write down the cycle indices of considered species $$Ens_0^4(A), \ Ens_0^5(O), \ Ens_0^6(B)$$

$$Z(A) = 1 + a_1 + {1 \over 2!}(a_1^2 +a_2) +... {1 \over 4!}(a_1^4+...+6a_4) $$

$$Z(B) = 1 + b_1 + {1 \over 2!}(b_1^2 +b_2) +... {1 \over 5!}(b_1^5+...+24b_5) $$

$$Z(C) = 1 + c_1 + {1 \over 2!}(c_1^2 +c_2) +... {1 \over 6!}(c_1^6+...+120c_6) $$

No further symmetric cycle indices are needed.

2) Then the cycle index of one non-empty basket is:

$$basket (a_1,a_2,a_3,a_4;\ b_1,b_2,...,b_5;\ c_1,c_2,...,c_6) = Z(A)Z(B)Z(C) - 1$$

3) in order to make a "plethystic substitution" we need the ingredients :

$$basket _2 = basket \ (a_2, a_4, 0, 0 \ ; \ b_2, b_4, 0, 0, 0 \ ; \  c_2, c_4, c_6,0,0,0)$$

$$basket _3 = basket \ (a_3, 0, 0, 0 \ ; \ b_3, 0, 0, 0, 0 \ ; \  c_3, c_6, 0,0,0,0)$$

$$basket _4 = basket \ (a_4, 0, 0, 0 \ ; \ b_4, 0, 0, 0, 0 \ ; \  c_4, 0, 0,0,0,0)$$

4) In the cycle index of $$Ens_4$$

$$ {1 \over 24}(X_1^4 + 6X_1^2X_2 + 3X_2^2 + 8X_1X_3+ 6X_4)$$

we replace X1, \ X2, \ X3, \ X4 \ with $$ \ basket, \ basket _2, \  basket _3, \ basket _4$$

5) now we can extract information from the cycle index Z just obtained.

$$Z^{types} = Z(a, a^2, a^3, a^4; \ o,o^2,...,o^5; \ b,b^2,...,b^6) $$

is the type indicator polynomial. The coefficient of $$a^4o^5b^6$$ in $$Z^{types}$$ is $$5808$$ and it represents the number of types of partitioning the 15 mixed fruits in four nonempty baskets.

6) By repeating the above for 3,2, and one nonempty bucket we get

$$ 5808 + 1383 + 104 + 1 = 7296 $$

7)

$$Z^{egf} = Z(a, 0, 0,0; \ o,0,...,0; \ b,0,...0) $$

is the e.g.f polynomial. The coefficient of $${a^4\over 4!}{o^5\over 5!}{b^6\over 6!} = 42355950$$  in Z^{egf} represents the different distributions of 4+5+6 labelled fruits in four baskets.

8) by considering the other cases of non-empty baskets, one has :

$$ 42355950 + 2375101 + 16383 + 1 = 44747435$$

9) we may do some double check. For example, 1+104 = 105 means to distribute 5 oranges in two baskets in three ways : 5+0, 4+1, 3+2; then the apples and bananas multiplies the number of cases to 3.5.7= 105.

10) the $$42355950$$ may be obtained faster by taking the coefficient of $$x^{15}$$ in

$$ {1\over4!} (x + {1\over 2!}x^2 + \cdots + {1\over 15!}x^{15})^4 $$

Indeed, when distributing labelled fruits, the colouring does not matter. Whatever comnposition of a total of 15 labelled apples, oranges and bananas will produce the same number of cases.