Let $\left(\delta_{i,j}\right)_{5\times5}$ be a matrix with $\delta_{i,i}=1$ and $\delta_{i,j}=\delta_{j,i}=1$ or $0$ when $i\ne j$ and $\delta_{i}=\sum_{j=1}^{5}\delta_{i,j}$. Let $$ f(k):=\max_{\delta_{1}+\delta_{2}+\delta_{3}+\delta_{4}+\delta_{5}=k}\delta_{1}\delta_{2}\delta_{3}\delta_{4}\delta_{5} $$ for $5\le k\le25$, then what is the combinatorial expression of $f(k)$ as a function of $k$?
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Welcome to MSE. Please include your own thoughts and the effort made thus far, so that people can work with you accordingly. (Please add those in the body of the question instead of commenting.) – Lee David Chung Lin Feb 25 '20 at 15:19
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Your $\delta_i$ is just how many of $1..5$ are equal to $i$, i.e., 1 if $1 \le i \le 5$. The maximum of the products of $\delta_i$ is thus one (if all 5 give one), and that can happen only if their sum is 5 (five of those can't ever sum more than 5). Thus $f(n) = \delta_{n, 5}$.
vonbrand
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It should be minimum in the definition of f(k), and had been corrected before you posted your answer. Thanks. – EmmaT Feb 25 '20 at 14:01
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Ever if for maximum, isn't the product with more 1's off the diagonal larger? – EmmaT Feb 25 '20 at 14:13
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What you posted does not answer the question. Could you please remove it? Thanks. – EmmaT Feb 25 '20 at 14:25