In how many ways can we distribute $11$ oranges and $6$ pears to $3$ kids, such that each kid gets at least $3$ orange and a maximum of $2$ pears?

Question

In how many ways can we distribute $11$ oranges and $6$ pears to $3$ kids, such that each kid gets at least $3$ orange and a maximum of $2$ pears?

I guess the generating function for the distribution of oranges is $(x^3+x^4+x^5)^3$. It doesn't have $x^6$ because if one kid had six oranges, then $11-6=5$ and one kid would have less than $3$ oranges. Then eash kid must have at least $3$, with a total of $9$. Then we must find the value of the coefficients from $x^9,x^{10},x^{11}$ on $(x^3+x^4+x^5)^3$.

Writing it as:

$$(x^3+x^4+x^5)^3=x^9(1+x+x^2)^3$$

Then we have to find the coefficients of $x^{9-9},x^{10-9},x^{11-9}=x^{0},x^{1},x^{2}$ in $(1+x+x^2)^3$. Rewriting it:

$$(1+x+x^2)^3=\left[\frac{1-x^3}{1-x}\right]^3=(1-x^3)^3 \left[ \frac{1}{1-x}\right]^3$$

Then:

$$(1-x^3)^3=\sum_{i=0}^3(-1)^i{3\choose i}x^{3i}\tag{A}$$

And:

$$\left[\frac{1}{1-x}\right]^3=\sum_{i=0}^{\infty}{3+i-1 \choose i}x^i\tag{B}$$

Then the coefficients for this part of the problem should be:

$$A_0B_0+A_0B_1+A_0B_2={3\choose 0}{2\choose 0}+{3\choose 0}{3\choose 1}+{3\choose 0}{4\choose 2}=1+3+6=10$$

And this number is bigger than the actual answer. I was planning to multiply this number by the result of the generating function for the pears, but I guess it's going to be bigger and farther from the answer. I tried to combine both generating functions, but it seems to cause a big hurdle to computation: $(1+ax+a^2x^2)^3 (b^3 x^3 + b^4 x^4 + b^5 x^5)^3$

Perhaps I am mistaken, but I don't know how to proceed to calculate this with generating functions.

Answer 1

If each of the three children receives a maximum of two of the six pears, then each child must receive two pears. If each of the three children receives at least three oranges, then there are two remaining oranges to distribute among the three children. This is equivalent to solving the equation $x_1 + x_2 + x_3 = 2$ in the non-negative integers, which can be done in $\binom{4}{2} = 6$ ways since it is the number of ways we can place two addition signs in a list of two ones.

Answer 2

First of all we give each kid $3$ oranges. Now there are $2$ oranges left to potentially distribute. Distributing $0$ oranges can be done only in $1$ way. Distributing $1$ oranges gives us $3$ choices and distributing $2$ oranges gives us $6$ ($2$ for one kid or a pair). So the total number of ways for the oranges is $10$. Now each kid gets either $0$, $1$ or $2$ pears. So we have $3^3=27$ ways for the pears. Now the total number of ways is $10*27=270 $.

Edit: If we have to give away all the fruit then each kid gets $2$ pears and we can distribute the remaining $2$ oranges in $6$ ways, as said before. So in this case, the final answer should be $6$.

Answer 3

For a generating function approach, consider the oranges and pears separately.

The OPSGF for the distribution of oranges is $$ f(x) = (x^3 + x^4 + x^5 + \dots)^3$$ You want the coefficient of $x^{11}$. It's true that we could disregard powers of $x^6$ or higher, but I think you will find the math simpler if we leave them in.

The series for the distribution of pears is $$ g(x) = (1 + x + x^2)^3$$ You want the coefficient of $x^6$.

Having found the solution to each of these sub-problems, multiply them together to find the number of ways to distribute both oranges and pears.