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1 vote

Any two disjoint subsets in a family of subsets intersect, prove that any maximal such family of subsets must contain $2^{n-1}$ subsets

You seem to be on the right track, so I'll just give you a hint: For any set $A$, you can prove that either $A$ or its complement $B=X\setminus A$ must be in $\cal{F}$. I'll fill in the rest of the ...
Einar Rødland's user avatar
1 vote

Number of paths to go from (0,0) to (6,4), each step can only be one up or to the right, and no three consecutive steps in the same direction?

Probably, there is a simpler solution, but it is the first one, which I came up with: there is a bijection between any way you can get from $(0;0)$ to $(6;4)$ and between number of ways to arrange $4$ ...
Vlad Boiko's user avatar

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