Combinatorics - How many ways to partition an integer n into k bins with values 0 - 5 and restrictions

Question

I have been struggling with this brain teaser for some time now. I looked at some combinatorics and partition equations but I can't find the one that captures the solution entirely.

Frame

I have a set with 6 elements (bins) { a , b , c , d , e , f }.

Each element in the set can take an integer value in the range (0,5).

The value N is given by N = a + b + c + d + e + f: The total range for N is (0,30).

Question

For each value of N in the range (0,30) how many unique combinations of elements can I have in the set without repetition.

e.g. if N = 30 there is only 1 possible unique combination {5,5,5,5,5,5}

also for N = 0 there is only 1 possible unique combination {0,0,0,0,0,0}

 for     N =  1 there are 6 possible unique combinations {1,0,0,0,0,0}
                                                         {0,1,0,0,0,0}
                                                         {0,0,1,0,0,0}
                                                         {0,0,0,1,0,0}
                                                         {0,0,0,0,1,0}
                                                         {0,0,0,0,0,1}

Therefore N can be described as a discrete normally distributed integer random variable in the range (0,30).

Please help.

If you are familiar with generating functions, you want the coefficient of $N$ in the expansion of $(1 + x + x^2 + x^3 + x^4 + x^5)^6$. — N. F. Taussig, Dec 16 '18 at 22:13
This is equivalent to asking how many ways there are to roll $n$ with $k$ dice, see https://math.stackexchange.com/questions/970523/how-many-ways-can-the-average-of-n-dice-be-a/970813#970813 — Mike Earnest, Dec 16 '18 at 22:17

Cheong Sik Feng · Answer 1 · 2018-12-17T17:25:06.443

You could as suggested by N. F. Taussig use the generating function $(1 + x + x^2 + x^3 + x^4 + x^5)^6$ and the answer will be the coefficient of $x^N$ of the expanded function.

Otherwise This question can be solved using the technique described here

Let there be $N$ stars. Since you want to separate them into $6$ bins, you will need to use $5$ bars. The $5$ bars will "split" the balls into 6 bins. The permutation will always be of the form $$\{(bin~a)|(bin~b)|(bin~c)|(bin~d)|(bin~e)|(bin~f)\}$$ Any permutation of the $N$ stars and $5$ bars will result in a unique solution. The number of ways that it can be arranged where $a, b, c, d, e$ and $f$ non-negative is $N+5\choose 5$.

However, this will include cases where the bins can be more than 5. We will need to exclude the cases where $a, b, c, d, e$ or $f$ is greater than 5. This can be achieved by the Principle of Inclusion-Exclusion. We would first subtract the case where only one of $a, b, c, d, e$ or $f$ is greater than 5. This is equivalent to doing the above but with $N-6$ instead of $N$. There are ${6\choose 1}$ ways for a bin to be more than 5. Hence you would subtract ${6\choose 1}{N-6\choose 5}$ from the total number of ways. However in this you will also notice that the case where 2 of the bins is greater than 5 was included twice. You would repeat the steps until the number of ways to have more than 5 balls in $j$ bins become zero.

The general formula would hence be $$\sum_{j=0}^{floor(\frac{N}{6})} (-1)^j\binom{6}{j}\binom{N+5-6j}{5}$$.

Be careful. Your solution includes those cases in which there are more than five stars in a bin. — N. F. Taussig, Dec 17 '18 at 10:24
Still .. another correction to do : the limit of summation is $\left\lfloor N/6 \right\rfloor $ and not $\left\lfloor (N+5)/6 \right\rfloor $: See my answer above. — G Cab, Dec 17 '18 at 16:57
Thanks for the correction. forgot that $N+5-6j >= 5$ not $0$ — Cheong Sik Feng, Dec 17 '18 at 17:26

G Cab · Answer 2 · 2018-12-17T14:17:04.127

Your problem in general reads as $$N_{\,b} (s,r,m) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered} 0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\ x_{\,1} + x_{\,2} + \cdots + x_{\,m} = s \hfill \\ \end{gathered} \right.$$ where $N_b$ is expressed through the sum $$ N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)} {\left( { - 1} \right)^k \binom{m}{k} \binom { s + m - 1 - k\left( {r + 1} \right) } { s - k\left( {r + 1} \right)}\ } $$ and whose o.g.f. in $s$ is as already suggested by N.F. Taussig $$ F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqslant \,\,r\,m} \right)} {N_b (s,r,m)\;x^{\,s} } = \left( {1 + x + \cdots + x^{\,r} } \right)^m = \left( {\frac{{1 - x^{\,r + 1} }}{{1 - x}}} \right)^m $$

The above is thoroughly explained in this related post and in this paper.

In your case, you just have to put $m=6,\;r=5,\; s=N$.

Thank you very much. This solves it perfectly. I can use this for analysis immediately although it will take further time to understand it. — Emmanuel Uwechue, Dec 17 '18 at 15:11
@EmmanuelUwechue: glad it helps. A further explanation of how it is built is the "inclusion-exclusion" principle described by S.F. Cheong, which is better grasped if you read the "geometric" interpretation given in the other cited post. — G Cab, Dec 17 '18 at 16:51

Combinatorics - How many ways to partition an integer n into k bins with values 0 - 5 and restrictions

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