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 Mar 4 comment Have 52 regular deck of cards, probability getting the first red ace? So would 52 choose 2 be the possible placements of red aces in the deck? Mar 4 comment Have 52 regular deck of cards, probability getting the first red ace? What does the (52-1-m)/(52-m+1) stand for? I understand that the second part would be the probability for the two red aces. Also, how did you find that the maximal is k=1? And thank you for your help!! Mar 3 comment Compositions with first part 1 Thank you!! What if the second part is 1 instead of the first part? Would it still be the same solution? Feb 25 comment Finding the transition probability matrix, two switches either on or off.. Thank you, you cleared this up extremely well for me. I can't thank you enough! I can finally go to bed :) Dec 12 comment Strong inducti0n with 3- and 5-peso notes and can pay any number greater than 7. Ahh it all makes sense now! Induction gets me every single time.. Dec 11 comment Strong inducti0n with 3- and 5-peso notes and can pay any number greater than 7. So we can assume that every integer smaller than n but greater than 8, can be expressed as a combination of 3- and 5- pesos. My confusion lies in proving n+3.. Dec 9 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. Dec 9 comment Choosing a 5 member team out of 12 girls and 10 boys Thank you. I didn't think it would be that simple. Dec 9 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! Dec 9 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? Dec 9 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? Dec 9 comment Having a forest and making it into a tree? Thanks again! I really appreciate your help! Dec 9 comment Having a forest and making it into a tree? Lol, thanks Brian! One question though... How did you know that m=10? Dec 8 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.. Dec 8 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. Dec 8 comment Planar Graphs with at least $2$ vertices and degrees at most $5$ So would the at most one exception be less than 6?