# MistyD

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bio website location age member for 11 months seen Sep 26 '12 at 13:03 profile views 171

# 313 Actions

 Aug26 comment Periodic Function - Repeating Pattern ProblemYeah. I just edited my last comment. Yeah same pattern Aug26 comment Periodic Function - Repeating Pattern ProblemIs their any way by which we could tell whether the $40$th bead would be Green,Yellow or Blue ? Is my method flawed ? Aug26 asked Periodic Function - Repeating Pattern Problem Aug25 comment If $x$ and $y$ are prime , which of the following cannot be their sum@GerryMyerson WOW. Thanks for the interesting fact!! Aug25 accepted If $x$ and $y$ are prime , which of the following cannot be their sum Aug25 comment If $x$ and $y$ are prime , which of the following cannot be their sumthats what I did. So I guess the only way is a hit and trial method Aug25 comment If $x$ and $y$ are prime , which of the following cannot be their sumThanks , so I could get an odd or an even result. Now am I suppose to try different combinations here of prime numbers ? Is there another way ? I mean I could go like $2+3 , 7+2 , 11+2 , ,13+3$ Aug25 asked If $x$ and $y$ are prime , which of the following cannot be their sum Aug18 asked In need of tips/suggestions when to add or multiply probabilities Aug18 comment Probability of getting $1$ out of $3$ shots@RobertIsrael its actually atleast Aug17 asked Probability of getting $1$ out of $3$ shots Aug16 accepted Combinatorics - How many committees are possible in this case Aug16 comment Combinatorics - How many committees are possible in this case@BrianM.Scott Thanks for clearing that up Aug16 comment Combinatorics - How many committees are possible in this case@GerryMyerson To get 6 , I grouped the two enemies as one so 7-1 = 6 and 4 was the committee requirement. Aug16 comment Combinatorics - How many committees are possible in this caseYou stated "To form an unacceptable committee, you must first choose the two enemies and then fill out the rest of the committee with any 2 of the remaining 5 people, so there are $_5C_2$ unacceptable committees." Could you explain this a bit more I still don't get it. How did you get 5 and how did you get 2 ? Aug16 asked Combinatorics - How many committees are possible in this case Aug7 accepted Need to clarify the “At-least Concept” in Combination. Aug7 comment Combination - Inverse Way of solving this problem@ladaghini I believe you are mistaking.. I have it correct for at-least one math teacher. Since All Possible Combinations - All possible English combination gives combinations that have at least one math in them Aug7 accepted Combination - Inverse Way of solving this problem Aug7 comment Combination - Inverse Way of solving this problem@joriki. I tried using that approach $\binom{9}{5} - ( \binom{5}{5} + \binom{6}{5} + \binom{7}{5} )$ but I dont get the correct answer