Using elementary combinatorics to solve dinner dates.

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"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.