*Originally posted by forkedknight*

**At first glance, I thought it was a trivially simple "yes".
**

After thinking about it some more, I don't see a way to construct groups that conflict.

If there were more than k groups, it would certainly be possible to create a situation were each person couldn't have their own chair.

I tend to agree. I wonder if it is possible to construct a proof by iteration, or some kind of diagonal argument. For a proof by iteration the cases k = 0, 1, 2 are trivial. In case 2 we either have committee 1 with {a, b} in and committee 2 with {c, d} in, or we have committee 2 with {a, c} in.

Doing the first step of the iteration by hand:

Take the two people committees {a, b} and {a, c} and add a person to each committee, we can't add a, b or c to either of these committees as either they are already on the committee or they're in the other one and a is already in both, call the new people d and e - they have to be different people again because a is already in both committees:

This gives us:

{a, b, d} and {a, c, e}

If committee 3 has a in then we need two more people f and g as a would prevent anyone else already on a committee from being on committee 3. Suppose instead we have b on committee 3, then both a and d are ruled out of being on that committee and we can either add two new person or one and only one of c or e to the committee and a new person, suppose we add e and f. Our three committees written in seating order are then:

{a, b, d}

{a, c, e}

{f, b, e}

Iterating by hand again, the case where a is in all three committees from the previous step is easy, the remaining committee members cannot be in any of the other committees. So starting from the less trivial case (and relabelling to make it less obscure):

{a, b, c, d}

{a, e, f, g}

{h, b, f, i} - again I've written them in seating order

There are a lot of possibilities for committee 4, but I don't think there's a conflict. The problem is I don't see a systematic way of generating a proof by iteration.