How many dinners did I have?

I figured this out by paying attention to, not the question, but the story. The story is a pattern. Day one; 6, day two; 5, day three; 4, day four; 3, day five; 2, day six; 1, and finally, day seven would be 0, just me all by myself. For every day spent, one less person attends the dinner.

Let us call the friends a, b, c, d, e and f.

Since I dinned with all the six on exactly once day, therefore at exactly one dinner I had all the six friends.

Since, I dinned with every five of them on two days, I must have had six dinners in which I dinned with

b c d e f ,

a c d e f ,

a b d e f ,

a b c e f;,

a b c d f,

and a b c d e; respectively

The above seven (1+6) dinners are such that every four of the friends were there in exactly 3 of these, and every there of them were There is exactly 4 of there.

Two of the friends say a and b, were present at the first dinner, and at four other dinners, that it at five dinners. Like wise every other pair of the friends was present at five of the above dinners.

Each of the friends were present at the first dinner and five others that is at six dinners.

Since each one of the friends was present at seven dinners in all, I must have had me dinner exclusively with each me of the six friends. It is clear that each one of the friends was absent at five of these six dinners, and therefore present at seven dinners in all, and absent at six out the thirteen dinners considered so far.

Since each friend was absent at seven dinners, therefore I must have had one dinner alone.