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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 11 months |
| seen | Sep 26 '12 at 13:03 | |
| stats | profile views | 171 |
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Aug 26 |
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Periodic Function - Repeating Pattern Problem Yeah. I just edited my last comment. Yeah same pattern |
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Aug 26 |
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Periodic Function - Repeating Pattern Problem Is their any way by which we could tell whether the $40$th bead would be Green,Yellow or Blue ? Is my method flawed ? |
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Aug 26 |
asked | Periodic Function - Repeating Pattern Problem |
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Aug 25 |
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If $x$ and $y$ are prime , which of the following cannot be their sum @GerryMyerson WOW. Thanks for the interesting fact!! |
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Aug 25 |
accepted | If $x$ and $y$ are prime , which of the following cannot be their sum |
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Aug 25 |
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If $x$ and $y$ are prime , which of the following cannot be their sum thats what I did. So I guess the only way is a hit and trial method |
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Aug 25 |
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If $x$ and $y$ are prime , which of the following cannot be their sum Thanks , 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 $ |
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Aug 25 |
asked | If $x$ and $y$ are prime , which of the following cannot be their sum |
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Aug 18 |
asked | In need of tips/suggestions when to add or multiply probabilities |
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Aug 18 |
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Probability of getting $1$ out of $3$ shots @RobertIsrael its actually atleast |
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Aug 17 |
asked | Probability of getting $1$ out of $3$ shots |
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Aug 16 |
accepted | Combinatorics - How many committees are possible in this case |
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Aug 16 |
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Combinatorics - How many committees are possible in this case @BrianM.Scott Thanks for clearing that up |
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Aug 16 |
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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. |
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Aug 16 |
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Combinatorics - How many committees are possible in this case You 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 ? |
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Aug 16 |
asked | Combinatorics - How many committees are possible in this case |
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Aug 7 |
accepted | Need to clarify the “At-least Concept” in Combination. |
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Aug 7 |
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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 |
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Aug 7 |
accepted | Combination - Inverse Way of solving this problem |
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Aug 7 |
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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 |
