Ways to put $50$ balls into $6$ distinct urns with an odd number in each urn?

Question

Say there are $50$ identical balls. We have $6$ different containers (note that these aren't named or anything, simply just distinct). How many ways can we put these balls into the urns so that each urn has an odd number? Okay. So, you have $23$ odd numbers from $1-50$ and six different urns. This leads you to say $\binom{23}{6}$, though that will not work due to the fact that these numbers you're selecting have a property: They must add up to $50$. How would I go about adding this property to my equation? How would I make an equation based on this? What kind of formula/theory is this question asking for? What am I missing in my equation or thought process?

Answer 1

These types of problems are in the field of generating functions. Generating functions are very useful tool to solve combinatorics problems using power series. I am putting here a link of very famous generating function book Link of the book. Moreover , you can find basic level explanation and applicaton of it in Kenneth Rosens Discrete mathematics and applications book Basic level book for generating functions

Anyway , i assume that you learned the technique , so i will start to solve. It is said that the boxes are distinct and contain odd number of balls where the balls are identical. Then , we can conclude that any box can contain such number of balls : $1,3,5,7,9,....,43,45$

As you see , we stop at $45$ because if each box contain odd number ,then a box can have at least $1$ balls. However , when we write the generating function form , do not need to restrict our generating function at $45$ in this question. (If you want you can restrict ,but it will make the process cumbersome )

So , the generating function form of an any box will be $$x^1 + x^3 +x^5 +x^7+....=\frac{x}{1-x^2}$$

Remember your calculus course to understand the meaning of $\frac{x}{1-x^2}$ , becasue $\frac{1}{1-x^2}= 1 +x^2 +x^4 +x^6 +.....$ , then $\frac{x}{1-x^2}=x(1 +x^2 +x^4 +x^6 +...)=x+x^3+x^5 +x^7 +...$

Moreover , realize the exponential of power series. The exponentials represents the desired number of balls that a box can contain.

Now , because of each boxes have the same restriction and there are $6$ boxes , we should find the expansion of the generating function. After that , we should find the coefficent of $x^{50}$ in this expansion such that $$[x^{50}]\bigg(\frac{x}{1-x^2}\bigg)^6$$

You can find the result by hand force using some techniques in given books , but it is cumbersome process (but not hard ). Hence , i would recommend you to use any software. I used wolfram-alpha such that Calculation of expansion

As you see in given link , the result is $80730$

Answer 2

Let $x_i$ be the amount of balls in the $i$th urn. We can rewrite our problem as

$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 50.$$

The $x_i$ have to be odd however, so each $x_i = 2y_i +1$, with no condition on the $y_i$ except that they are nonnegative integer numbers (so including zero). Our equation above becomes (after rewriting)

$$y_1 + \ldots + y_6 = \frac{50-6\cdot 1}{2} = 22.$$

We can count the number of ways to write a sum of 6 numbers which are equal to 22 by using the stars and bars methode. We 'separate' the numbers using $6-1 = 5$ bars and need to add 22 stars (representing the amount of units in each number). There are now a total of 5 + 22 = 27 positions to fill. An example: $22 = 4 + 1 + 8 + 0 + 9 + 0$ would be represented as $$****|*|********||*********||$$

Our counting problem has turned into a counting problem where we count in how many ways we can place 22 stars into 27 positions. Repitition is not possible (each position can only be filled once) and order does not matter (placing a star in the first and then placing one in the second position or vice versa results in the first and second position being filled in).

The number of ways to do this equals $\binom{22 + 6 -1}{22}=\binom{27}{22} = 80730$.