Sovardipk

Answered

2022-06-29

Monomials with degree k of the following polynomial $(1+{\mathit{x}}_{1}+\cdots +{\mathit{x}}_{1}^{q}+{y}_{1}+\cdots +{y}_{1}^{d}{)}^{n}$

Answer & Explanation

Dayana Zuniga

Expert

2022-06-30Added 16 answers

Move s and s′ as you wish as

$\sum _{s=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{s}{\textstyle )}\sum _{{s}^{\prime}=0}^{k}(\text{\# ways}{x}_{i}\text{add to s'})\cdot (\text{\# ways}{y}_{j}\text{add to k-s'}),$

using stars and bars(and inclusion-exclusion), we know that the number of tuples adding to s′ with parts less or equal than q is

$\sum _{r=0}^{s}(-1{)}^{r}{\textstyle (}\genfrac{}{}{0ex}{}{s}{r}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{{s}^{\prime}-(q+1)r+s-1}{s-1}{\textstyle )},$

similarly for the ${y}_{j}$ notice that we are not allowing 0 as a part (I took $1={x}_{i}^{0}$) and we have

$\sum _{\ell =0}^{n-s}(-1{)}^{\ell}{\textstyle (}\genfrac{}{}{0ex}{}{n-s}{\ell}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{k-{s}^{\prime}-d\ell -1}{n-s-1}{\textstyle )}.$

Plugging all together, we get

$\sum _{s=0}^{n}\sum _{{s}^{\prime}=0}^{k}\sum _{r=0}^{s}\sum _{\ell =0}^{n-s}(-1{)}^{r+\ell}{\textstyle (}\genfrac{}{}{0ex}{}{n}{s}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{s}{r}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{{s}^{\prime}-(q+1)r+s-1}{s-1}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{n-s}{\ell}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{k-{s}^{\prime}-d\ell -1}{n-s-1}{\textstyle )}$

Not entirely sure if this sum simplifies.

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