Q:

Combinatorics - too much combinations ? where is my mistake?

Problem : 

A person has 7 friends . How many combinations exists so that he would be able to  invite different
    groups of 3 of them for a dinner that lasts 7 days  (7
    sequential days) so that every 2 friends would be together in only
    one dinner ?

My solution : If a person has 7 friends then the number of couples that could be are 7 above 2 : $\binom{7}{2}$ , which are 21.Since every couple of friends are together in exactly one dinner (out of the 7 dinners in total) , then in every day that he invites 3 different friends , he actually invites 3 different couples , that don't appear in the rest of the days .Hence , in the first day he needs to pick 3 couples out of 21 , which are : $\binom{21}{3}$ . In the second day he needs to pick 3 couples out of 18 ... and so on . 
Then in total : 
$ \binom{21}{3} \cdot \binom{18}{3} \cdot \binom{15}{3} \cdot \binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3} \cdot \binom{3}{3} $
But this seems wrong , since the number of combinations is too big . Can anyone verify that I'm wrong ? 
Much appreciated 

A:

Gregory's answer explains what is wrong with your method of counting. Here's a way to get the right result:
First, since there are 21 pairs of friends and each of the 7 dinners "uses up" 3 pairs of friends, every pair of friends needs to occur in one of the dinner. Thus every friend will be invited to exactly three dinners and meet each of the 6 other friends once.
We start by counting how many different sets of 7 dinners there are, without caring about which sequence the dinners are held in. Usually this tends to be harder than counting ordered sequences, but in this particular problem it happens to be much easier.
Let A be the youngest friend. The other six friends needs to go to three dinners with A in them in three pairs. The number of ways to partition six elements into three unordered pairs is $5\cdot3\cdot1=15$ (see e.g. Brian M. Scott's answer here), so there are 15 ways to choose what the three dinners that include A will be.
Let B be the youngest friend except for A, and let C be the third friend at the dinner that A and B are in. Let D be the youngest friend not yet named, and let E be the third quest at the AD dinner. The dinners decided on so far are now:
ABC
ADE
AFG

There must be a dinner with B and D in it, and the third person there is either F or G. There are 2 options to choose among here.
Without loss of generality let's assume it was F, so our dinner list now looks like
ABC
ADE
AFG
BDF

At this point there's one and only one way to arrange the three remaining dinners. The BE diner can only be BEG, and then the CF and CE dinners must be CDG and ORY CEF.
Thus, multiplying together, the number of ways to select an unordered set of 7 dinners is $15\cdot 2=30$.
It remains to select which order the 7 dinners will be held in. There are $7!=5040$ ways to do that, so the final number of options is
$$ ((5\cdot 3\cdot 1)\cdot 2)\cdot7! = 30\cdot 5040 = 151\,200$$

How I came up with this: I started doing it "the hard way", caring about the sequence of dinners from the beginning. It took about one page of paper to exhaustively enumerate the possible shapes of the who-has-met-whom-yet graph after each night. Then I spent a few hours trying to condense the big exhaustive diagram down to a proof. Along the way, a gap in one of my proofs turned out to be because the lemma I was trying to prove wasn't true, because I had made an error during the exhaustive search, which I then needed to fix. Finally I came upon the above argument. All in all an interesting interplay and cross-fertilization between brute-forcing through a concrete examples and the desire to find an abstract argument.

