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 Dec9 comment Choosing a 5 member team out of 12 girls and 10 boys I haven't thought about doing it that way! Brian, you have been helping me a lot, and I really appreciate that someone with your knowledge is on here helping others out. Thank you. Dec9 comment Choosing a 5 member team out of 12 girls and 10 boys Thank you. I didn't think it would be that simple. Dec9 asked Choosing a 5 member team out of 12 girls and 10 boys Dec9 accepted Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. Dec9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. OH! So there are 1001 pigeonholes and 1024 pigeons! Got it! Thank you Brian! Dec9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. @BrianM.Scott, ooh alright, so the 1024 subsets are the pigeons, but what would be the pigeonholes? All the possible sums? Dec9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. So the max sum of ages is 1000 ( ten people and they all can be 100 yrs. old). And the number of subsets is 2^(10) - 1? Dec9 asked Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. Dec9 comment Having a forest and making it into a tree? Thanks again! I really appreciate your help! Dec9 awarded Supporter Dec9 accepted Having a forest and making it into a tree? Dec9 comment Having a forest and making it into a tree? Lol, thanks Brian! One question though... How did you know that m=10? Dec8 comment Having a forest and making it into a tree? Well, we can have 100 trees (just a dot) in a forest at most, but we also have 90 edges.. So if we add 1 edge to each of those trees, then we will have 50 trees in the forest, but we will have 50 edges.. Dec8 asked Having a forest and making it into a tree? Dec8 comment Planar Graphs with at least $2$ vertices and degrees at most $5$ Suppose every vertex, with at most one exception, has degree at least 6. Then, 2E =< 2(3V-6) = 6V-12; Sum (deg V) >= 6(V-1) = 6V-6; We see that 6V-12 > 6V-6, thus there are at lease two vertices whose degrees are at most 5. Is this the solution? I still can't see how we have two vertices that are at most 5. Dec8 comment Planar Graphs with at least $2$ vertices and degrees at most $5$ So would the at most one exception be less than 6? Dec7 awarded Scholar Dec7 accepted Planar Graphs with at least $2$ vertices and degrees at most $5$ Dec5 asked Planar Graphs with at least $2$ vertices and degrees at most $5$ Oct22 awarded Student