How Many Combinations of Two Days Can Be Chosen from Seven?

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To determine the optimal combination of two meeting days from a week, one can analyze the preferences of participants and their potential attendance. The discussion highlights the importance of balancing the average percentage attendance (Pa) with the percentage of individuals who can attend at least once (Po). By creating a table of combinations and plotting the results on a scatter plot, one can visualize the trade-offs between these metrics. The example provided indicates that there are 49 possible combinations for two days, which simplifies the analysis process. This method can be extended to three or four days, requiring more detailed programming and data organization.
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A certain number of persons (for example 100) choose the days of the week they prefer for a meeting, a fest, an event. The day preferred by most of the persons (for example 40) is Monday. If the meeting is every Monday, at the end of the year the number of persons who partecipated would be the greater, but the percentage of persons would be the same (40%) - for hypothesis, if one person does not indicate one day, he will never be able to partecipate on that day. How do you calculate the days to alternate to satisfy the greatest number of persons AND the biggest percentage of them?

(Every person can indicate one to seven days. We know of course the distribution of the votes)
 
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j0anna said:
... How do you calculate the days to alternate to satisfy the greatest number of persons AND the biggest percentage of them?

For small multi-objective combinatorial optimization problems like these, an easy strategy is to make a table of all possible combinations and plot the results in a scatter plot and choose the least objectionable alternative.

Say if you're looking to alternate two different days and want a tradeoff between Pa (average percentage attendance) and Po (percentage who can attend at least once), that's only 49 combinations and is fairly easy to implement in Excel or otherwise. If Tuesday has 30% vote and 50% can attend at least one of Monday or Tuesday, the table would begin like:

D1 D2 Pa Po
M M 40% 40%
M T 35% 50%
...

Obviously neither M-M or M-T wins on both measures, but doing a scatter plot of Pa vs Po will nicely visualize the relative merits of each combination and give you a small list of best candidates to choose from.

Same idea applies if 3 or 4 days are alternated, just a bigger table and more careful programming is necessary.

HTH
 
bpet said:
For small multi-objective combinatorial optimization problems like these, an easy strategy is to make a table of all possible combinations and plot the results in a scatter plot and choose the least objectionable alternative.

Say if you're looking to alternate two different days and want a tradeoff between Pa (average percentage attendance) and Po (percentage who can attend at least once), that's only 49 combinations
How do you obtain 49?
 
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