Water Combinatorics

The scenario at hand can be handled by using a generating function approach.

The sum of volumes of water in each container should be 46 L.

Also, volume of water is quantized (minimum $=1 \ \text{L}$ ).

You would look for the coefficient of $x^{46}$ in

$(1+x+x^2+x^3)^6(1+x+x^2+....+x^8)^4 \\ =\left(\displaystyle\sum_{k=0}^{3} x^k\right)^6 \left(\displaystyle\sum_{k=0}^{8} x^k\right)^4$

That would be equivalent to the coefficient of $x^4$ in the above polynomial, because the maximum power of $x$ in the polynomial is $50$ , and the polynomial is symmetric (product of two symmetric polynomials is symmetric).

So, the desired answer would be the coefficient of $x^4$ in $P=(1-x^4)^6(1-x^9)^4(1-x)^{-10} \ \text{(Sum of geometric series)}$

Now, the coefficient of $x^r$ in the Maclaurin series expansion of $(1-x)^{-10}$ is $\binom{10+r-1}{r}$ .

So, the coefficient of $x^4$ in $P$ can be found by first taking $x^4$ from the 1st bracket and $x^{0}$ from the 2nd and 3rd brackets, and then $x^0$ from the 1st and 2nd brackets and $x^4$ from the third bracket.

Therefore, desired answer is $\binom{10+4-1}{4}-\binom{6}{1}=\binom{13}{4}-6=\boxed{709}$

There may be other ways to solve this problem involving case analysis.

This problem can be solved rapidly using a different perspective on the problem.*

Between our 10 containers, we have the capacity to hold 50 liters. 46 liters are being distributed, so there will be a total of 4 liters of "empty space" within the ten containers.

Notice that each distribution of the 46 liters corresponds to a unique distribution of 4 "anti-liters" of water. We will count these instead.

By a simple application of stars and bars (or balls and urns), there are 13C4 ways to distribute the 4 anti-liters of water. However, a 3 liter container cannot contain 4 liters of space, so we must subtract away the 6 distributions we counted in which we allocated all 4 liters of anti-water to a single 3 liter container.

13C4 -6 = 715 - 6 = 709.

*This method is extremely effective but not always applicable; for example, had the problem specified the distribution of 36 L of water, this tactic would be far less useful.