# Using elementary combinatorics to solve dinner dates.

"Pankaj has six friends and doing a vacation he met them over dinners. He found that he dined with all 6 friends exactly 1 day, every 5 of them 2 days, every 4 of them 3 days, every 3 of them 4 days, and every 2 of them 5 days. Further every friend has been present at 8 dinners and been absent at 8 dinners.

Based on these, how many dinners did Pankaj have during the vacation, and how many of those were by himself?"

This is a problem that has been featured in some competitive tests in India, and also relevant preparatory material. [Directly quoted, so cannot provide further clarification] However no in depth explanation of the method of counting is present. I fail to see the method, and would strongly urge you to recommend a method of counting.

Edit:

Regarding Karn's second comment, I am enclosing the exact text of text of the question from one source, but if you search Google Books, you will realise that a lot of questions of this format exist, unfortunately all belonging to objective testing without elaborate solutions. Upon clicking here, please follow questions 7-9.

As per the comments of Jose and Ethan, I have included the working which I previously omitted because I thought it was not substantial.

As Karn's first comment states, that was my initial claim.

Since, for each dinner, every person can have only 2 decisions - either attending or not attending, then the sum of the two must be the total number of dinners. However, the question statement, gives:

$6 \choose 6$+$6 \choose 5$*2+$6 \choose 4$*3+$6 \choose 3$*4+$6 \choose 2$*5+$6\choose 1$*m+$6 \choose 0$*n = 16

where $m$ and $n$ are the number of times he has dined with each friend and the number of times he dined alone, respectively.

However, here, visibly, LHS > RHS which defeats our claim that the total number of dinners was 16.

This is how far I have progressed. But I can't see any errors in their respective reasoning.

• Welcome to stackexchange. You are more likely to get answers rather than downvotes and votes to close if you edit the question to show us what you tried and where you are stuck. – Ethan Bolker Mar 27 '18 at 16:18
• Wouldn't there be exactly 16 dinners in total if every friend is present for 8 and absent for 8? And he clearly has exactly 1 dinner by himself by simply counting the days where he has other dinners.... – Karn Watcharasupat Mar 27 '18 at 16:25
• Also there are 50 'dinner seats' that need to be filled but the 6 friends can only provide 6*8=48 'dinner seats' in total so is there any typo? – Karn Watcharasupat Mar 27 '18 at 16:58
• @JoséCarlosSantos, thanks for the welcome, unfortunately, I could progress only this much. – Wrik Karmakar Mar 27 '18 at 18:20
• @EthanBolker, thanks for the welcome, unfortunately, I could progress only this much. – Wrik Karmakar Mar 27 '18 at 18:20