my Combinatorics class followup

DPatrick

There were 3 problems from the worksheet that we didn't cover in class. Here are their answers and some additional hints.

#13. Instead of trying to count the number of anchors per set, try to count the number of sets per anchor. In other words, count the number of sets that have 1 as an anchor, the number of sets than have 2 as an anchor, and so on. (The final answer is that there are a total of $(n-2) \cdot 2^{n-3}$ anchors, so since there are $2^n$ sets there is an average of $(n-2)/8$ anchors per set.)

#17. The answer is "yes". The word "three" in the problem statement is the biggest clue. Another clue is that $2007 < 2^{11}$ and $3^{10} < 100000$ .

#18. Try proving it for 5 points first. (This first part is mostly geometry.) Then think about how your analysis of the 5-point case expands to a greater number of points. (This second part is all combinatorics.)

**Posted:** Wed Jul 29, 2009 10:45 am

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