Combination of flavors We want to make ice cream cones using some choices from 4 flavors, but with the following rules:
None of the 4 can be used in all 3 cones and, for any 2 cones, there is at least 1 flavor at both of them but not at the 3rd one. In how many ways can we make these 3 cones while satisfying the constraints?
Each cone can have from 0 up to 4 flavors but using the constraints, we deduce that we can only have 2 or 3 in each of them.
More specifically, we can't have 3 flavors at all 3 cones, we can only have 3+3+2 or 3+2+2 or 2+2+2 - right?
I have counted 6+24+4 ways in total, so 34 and this number x 6 for the rotation of the 3 cones (assuming they are distinct). Am I right?
 A: Your logic is entirely right, but I believe you've miscounted the ways of having 3+2+2 flavours.
Since each pair of cones must share one flavour that the third doesn't have, and there are ways of choosing a pair of cones, three of your flavours must be assigned in this way - e.g. flavour $A$ to cones $1$ and $2$, $B$ to cones $1$ and $3$, and $C$ to cones $2$ and $3$ - leaving one flavour, $D$.
As you rightly note, this leftover flavour can then either be placed on no cones, 1 cone, or 2 cones, but not all 3. 


*

*If we have 2+2+2, one flavour is left out, and permuting the other three is the same as permuting cones. There are thus $4!/3! = 4$ ways to assign flavours.

*If we have 3+2+2, e.g. $ABD$, $AC$, and $BC$, then one flavour only appears once, and on the 3-scoop cone ($D$) and one flavour is not on that cone ($C$). Swapping the other two flavours ($A$ and $B$) is the same as swapping the 2-scoop cones, so $A$ and $B$ are indistinguishable. So there are $4!/2! = 12$ ways of doing this. This is where we differ.

*If we have 3+3+2, e.g. $AB$, $ACD$, and $BCD$, then swapping $A$ and $B$ is equivalent to swapping the latter two cones, while $C$ and $D$ are clearly indistinguishable from each other. We have two pairs of indistinguishable flavours, so there are $4!/(2!*2!)=6$ ways of assigning flavours.


For each case we can then permute the cones, and there are as you say $6$ ways of doing this. I thus get a total of $(4+12+6)*6 = 22*6 = 132$.
