I need to find the sum of coefficients in expansion of
$$(x_1+x_2+x_3+x_4+x_5+x_6+x_7)^{11}$$ in which degree of any variable is not zero?
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2Sum of coefficients is the value of this polynomial when $x_1=x_2=x_3=x_4=x_5=x_6=x_7=1$. Recall Multinomial theorem. Also, what do you mean by saying that degree of any variable isn't zero? – Jakobian Aug 20 '18 at 08:56
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@Rumpelstiltskin: my guess is that the coefficients of monomials like $x_1^4 x_6^5 x_7^2$ have to be excluded from the count. – Jack D'Aurizio Aug 20 '18 at 17:57
2 Answers
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Supposing that we ask the question for $$\left(\sum_{q=1}^n x_q\right)^p$$ we find by inclusion-exclusion $$\sum_{k=0}^n {n\choose k} (-1)^k (n-k)^p$$ so that for $n=7$ and $p=11$ the answer becomes
$$322494480.$$
Alternatively we obtain by inspection the closed form
$$n! \times {p\brace n}.$$
We take the problem to mean that we sum the coefficients on all terms of the expanded product that contain all $p$ variables. The poset for PIE consists of nodes $Q\subseteq [n]$ representing those monomials of total degree $p$ where the variables $x_q$ with $q\in Q$ are missing, plus possibly more. The computation of the weights is the same as at this MSE link.
Marko Riedel
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would you please elaborate what you mean by "we find by inclusion-exclusion" , or you can suggest some source to understand the concept. Thanks in advance. – Gyan Chand Kewat Aug 22 '18 at 08:51
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We have $$(x_1+\cdots +x_n)^m=\sum_{m_1+\cdots m_n=m}x_1^{m_1}\cdots x_n^{m_n}$$therefore $$\text{The sum of the coefficients}=\sum_{m_1+\cdots m_n=m}1=n^m$$
Mostafa Ayaz
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