2

I have $(s+s^2+s^3+s^4+s^5+s^6)^7$, and I'm trying to find the coefficient on $s^{14}$. I've tried using the multinomial theorem, but that leads to the problem of finding all $k_1, k_2, \ldots , k_6$ such that $\sum_{n=1}^6 k_n = 7$ and $\sum_{n=1}^6nk_n = 14$, and that doesn't seem to put me any closer to an answer.

I tried rewriting it as $s^7(1+s+s^2+s^3+s^4+s^5)^7$ and looking for the coefficient of $s^7$ in the right half, $(1+s+s^2+s^3+s^4+s^5)^7$, but that doesn't make things much easier.

joriki
  • 238,052
Greg
  • 155

2 Answers2

5

This is equivalent to counting the number of ways of writing $14$ as an ordered sum of $7$ summands from $1$ to $6$, which is equal to the number of ways of distributing $14$ balls over $7$ non-empty bins with capacity $6$, which is equal to the number of ways of distributing $14-7=7$ balls over $7$ bins with capacity $5$, which is given by

$$ \binom{7+7-1}{7-1}-7\binom{7+7-1-6}{7-1}=\binom{13}6-7\cdot7=1716-49=1667\;, $$

where the second term subtracts the configurations counted by the first term that exceed one of the capacity restrictions (see also Balls In Bins With Limited Capacity).

joriki
  • 238,052
1

Binomial Series Approach

Using the Binomial Theorem and negative binomial coefficients, $$ \begin{align} \left(s+s^2+s^3+s^4+s^5+s^6\right)^7 &=s^7\left(\frac{1-s^6}{1-s}\right)^7\\ &=s^7\sum_{k=0}^7\binom{7}{k}\left(-s^6\right)^k\sum_{j=0}^\infty\binom{-7}{j}(-s)^j\\ &=s^7\sum_{k=0}^7(-1)^k\binom{7}{k}s^{6k}\sum_{j=0}^\infty\binom{j+6}{j}s^j\tag1 \end{align} $$ Using Cauchy Products, the coefficient of $s^n$ in $(1)$ is $$ \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}\tag2 $$ Plug $n=14$ into $(2)$: $$ \bbox[5px,border:2px solid #C0A000]{\binom{7}{0}\binom{13}{7}-\binom{7}{1}\binom{7}{1}=1667}\tag3 $$

The Sum in $\bf{(2)}$ represents a Polynomial

When $n\ge43$, $n-6k-1\ge0$ for all $k\le 7$. This means that $$ \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7} = \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{6} $$ which is an order $7$ repeated difference of a degree $6$ polynomial in $k$. Therefore, for $n\ge43$, $$ \sum_{k=0}^7(-1)^k\binom{7}{k}\binom{n-6k-1}{n-6k-7}=0 $$ That is, for $n\ge43$, the coefficient of $x^n$ vanishes.

robjohn
  • 345,667