Number of possible quintuples $(n_1, n_2, n_3, n_4, n_5)$?

Asked Jan 24 '22 at 23:08

Active Jan 24 '22 at 23:11

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Suppose there are $10$ students participating in a workshop that meets every day of a five-day work week. Each student must present a poster in exactly one of the days of the workshop, and every day of the workshop should have at least one poster presentation. If $n_k$ denotes the number of posters presented in day $k$ of the workshop, what is the number of possible quintuples $(n_1, n_2, n_3, n_4, n_5)$?

We can reformulate this problem as those tuples $(n_1, n_2, n_3, n_4, n_5)$ such that $\sum_{i=1}^5 n_i = 10$ and $n_i\geq 1$. Now I know that this integer program has $126$ solutions. Would this be the correct answer?

edited Jan 24 '22 at 23:11

N. F. Taussig

76,571

asked Jan 24 '22 at 23:08

chesslad

2,533

2

Yes, that is correct. Are you asking how to show that? – N. F. Taussig Jan 24 '22 at 23:12
1

In this answer, ignore the Inclusion-Exclusion theory and focus on the Stars and Bars theory. – user2661923 Jan 24 '22 at 23:22
@N.F.Taussig I only want to confirm if the reformulation is correct – chesslad Jan 24 '22 at 23:29
@user2661923 precisely how I solve it, but want to know if the reformulation is correct. That is, are the two problems the same or if I'm missing something – chesslad Jan 24 '22 at 23:30
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If you solved it the same way as that done in the linked answer, then yes, your approach was valid. That is, the set of solutions to $~x_1 + \cdots + x_k = n \in \Bbb{Z^+} ~: ~x_i \in \Bbb{Z_{\geq 1}}~$ bijects to the set of solutions to $~y_1 + \cdots + y_k = (n - k) ~: ~y_i \in \Bbb{Z_{\geq 0}}.$ – user2661923 Jan 24 '22 at 23:34
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Your reformulation is correct. – N. F. Taussig Jan 24 '22 at 23:41
@N.F.Taussig Thank you – chesslad Jan 24 '22 at 23:47

Number of possible quintuples $(n_1, n_2, n_3, n_4, n_5)$?

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