Suppose 51 women and 51 men are sitting at a round table. Prove that there exists a human (either a woman or a man) both of whose neighbours are women. (Hint: at least how many groups of women are sitting together?)
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To disprove the proposition we can attempt to block WMW or WWW arrangements, where W=woman and M=man. (In particular, a group of n women creates n-2 WWW arrangements.) Since this involves groups of 3 people, 102 people can be split into 34 groups of 3 people. If we use the circular arrangement WMM in a chain we get WMMWMMWMM... However, we need 34 such groups, and that means too many men (68) and too few women (34). The arrangement MWW in a chain (...MWWMWWMWW...) inhibits three women from sitting together, but creates WMW arrangements. These should be minimal, though.

If we use 17 groups of WMM and 17 of MWW we get 34+17=51 men and 17+34=51 women. This arrangement therefore gives us the required number of people, while creating 16 WMW arrangements. (The last WMM group is followed by MWW which creates the only MMM arrangement and reduces 17 WMW arrangements to 16.) This is the minimum number for 51 pairs of men and women, therefore the proposition is true. There is at least one example satisfying the proposition—in fact, the minimum is 16.

ago by Top Rated User (839k points)