# Number of ways to distribute items (with limit of min and max at each location)

There are a total of $n$ objects in stock. You must keep at least $c_i$ objects in stock at location $i$. Assume $n \ge \sum c_i$.

a.) How many different inventories $(x_1,x_2,\dots,x_r)$ are there?

b.) Each location can store a maximum of $d_i$. Let $r = 2$ and count the number of different invetories $(x_1,x_2)$ with $c_1 \le x_1 \le d_1$ and $c_2 \le x_2 \le d_2$.

My answer for part a: We have no choice over where $c_1 + c_2 + \dots + c_r = c_t$ objects are placed. The ramaining $n-c_t$ objects can be spread across $r$ locations. So, there are $$\binom{(n-c_t) + r - 1}{r-1}$$ different inventories.

My incomplete answer for part b: After placing minimum required objects at each location, we are left with $n-c_1-c_2$ objects to distribute.

The remaining objects can be distributed in $$\binom{(n-c_1-c_2)+2-1}{2-1} = (n-c_1-c_2)+1.$$

Now from the number above, I need to subtract the number of combinations that lead to overflows in the two locations.

[---Not sure if the rest is correct---]

Number of ways location 1 overflows:
Location 1 overflows when $x_1 > d_1$ and $c_2 \le x_2 \le d_2$: $$\binom{(n-(d_1+1)-c_2)+2-1}{2-1} = n-d_1-c_2$$

Number of ways location 2 overflows:
(same reasoning as above) $$n-d_2-c_1$$

Total number of inventories is $$(n-c_1-c_2+1)-[(n-d_1-c_2)+(n-d_2-c_1)] = (d_1+d_2+1)-n$$

Total number of inventories $= (d_1+d_2+1)-n$ doesn't make sense.

-

Your answer to part (a) is fine. In (b) you’ve overlooked the possibility of overflow in both inventories simultaneously.

You got off to a fine start: there are $$I_0 = n-c_1-c_2+1$$ inventories that meet the minimum requirements. Then you want get the number of inventories that overflow location $1$ but meet the minimum requirement at location $2$; your answer of $n-d_1-c_2$ is correct only if $n\ge d_1+c_1$. If $n < d_1+c_1$, it’s not possible to overflow at location $1$ and meet the minimum requirement at location $2$. Thus, the correct result for this step is $$I_1 = \max\{n-d_1-c_2,0\}$$ inventories that overflow at location $1$ and meet the minimum at location $2$. Similarly, there are $$I_2=\max\{n-d_2-c_1,0\}$$ inventories that meet the minimum at location $1$ but overflow location $2$.

However, there may be inventories that overflow at both locations. When you subtract off the inventories that overflow at location $1$ and then those that overflow at location $2$, the inventories that overflow at both locations get subtracted twice, so you have to add them back in.

The basic calculation gives $$\binom{n-(d_1+1)-(d_2+1)+2-1}{2-1} = n-d_1-d_2-1$$ of these inventories, but again this is true only if $n\ge d_1+d_2+1$; if $n < d_1+d_2+1$, it isn’t possible to overflow both locations. Thus, the correct figure is $$I_3=\max\{n-d_1-d_2-1,0\}.$$

The final result is then given by $$I_0-I_1-I_2+I_3.$$

Note that if $n \ge d_1+d_2+1$, this evaluates to $0$. This makes perfectly good sense: in that case it’s impossible to meet the requirements, and there are no acceptable inventories.

-

Unless I am reading the problem wrong, this is the same as a couple of other problems that were asked in the last couple of days. You are given a total and want to find the number of ways to make that total $\left(\displaystyle\sum_{i=1}^rx_i=n\right)$ given a range constraint on each summand $(c_i\le x_i\le d_i)$. The answer is the coefficient of $x^n$ in $$\prod_{i=1}^r\frac{x^{d_i+1}-x^{c_i}}{x-1}\tag{1}$$ We can apply this to the special cases you have listed.

a) you let $d_i\to\infty$. In that case, $(1)$ becomes $\displaystyle\prod_{i=1}^r\frac{x^{c_i}}{1-x}=x^{\sum_{i=1}^rc_i}\frac{1}{(1-x)^r}$, so the answer becomes the coefficient of $x^{n-\sum_{i=1}^rc_i}$ in $\frac{1}{(1-x)^r}$. \begin{align} \frac{1}{(1-x)^r} &=\sum_{i=0}^\infty\binom{-r}{i}(-x)^i\\ &=\sum_{i=0}^\infty\binom{r+i-1}{i}x^i\\ &=\sum_{i=0}^\infty\binom{r-1+i}{r-1}x^i \end{align} Therefore, the answer is $\displaystyle\binom{r-1+n-\sum_{i=1}^rc_i}{r-1}$.
b) you set $r=2$. In this case, $(1)$ becomes $$\left(x^{c_1}+x^{c_1+1}+\dots+x^{d_1}\right)\left(x^{c_2}+x^{c_2+1}+\dots+x^{d_2}\right)\tag{2}$$ and the answer is the coefficient of $x^n$ in $(2)$.

Suppose $d_1-c_1\le d_2-c_2$, then the answer is $$\text{the coefficient of }x^n\text{ in }(2)=\left\{\begin{array}{}n-c_1-c_2+1&\text{for }c_1+c_2\le n\le d_1+c_2\\d_1-c_1+1&\text{for }d_1+c_2+1\le n\le c_1+d_2-1\\d_1+d_2-n+1&\text{for }c_1+d_2\le n\le d_1+d_2\\0&\text{otherwise}\end{array}\right.$$

-
Please show the closed form, I can't get the answer. – jamesio Oct 3 '11 at 17:04
@jamesio: I have included the gory details. – robjohn Oct 3 '11 at 17:53