ovie8 Posted September 17, 2008 Share Posted September 17, 2008 Whoa. What school do you go to? Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted September 17, 2008 Share Posted September 17, 2008 ND Link to comment Share on other sites More sharing options...
MillerGenuineDraft Posted September 17, 2008 Share Posted September 17, 2008 It would be funny if someone asked for help in gym. :D Link to comment Share on other sites More sharing options...
canuckster19 Posted September 17, 2008 Share Posted September 17, 2008 not to be harsh, but almost every step here has a significant error in it... not trying to be rude but the guy who asked the question shouldn't use this as an example. to choose all possible combinations you don't take 12x11x10x9. This is counting permutations, not combinations. division by 2 does not compensate for this, to get combinations you have to divide by 4! = 4x3x2x1 = 24, which is the number of ways of arranging 4 unique objects among 4 people (not 16 as was stated). This division divides out the number of extra ways we would be able to arrange the prizes if they were unique. So, the number of ways of arranging the four identical prizes between four couples is 12(choose)4 = 12x11x10x9 / 4x3x2x1 = 495. Now that we have counted combinations properly, no other multiplication is necessary to deal with selecting the couples. In the post above, the division by 2 to get combinations should be a division by 24, and there should be no multiplication after. Such a multiplication would be to account for 4 unique prizes, which is not what we have here (and, 16 is simply wrong, and, if there were unique prizes, the best advice is to count permutations, not combinations, in the first step to avoid other possible errors such as what happened there with an incorrect calculation of the number of ways to distribute 4 items among 4 people). Now that the couples have been selected, there are two members of each couple. So for each of the 4 couples who get a prize, there are 2(choose)1 = 2 ways of them being awarded the prize. This factor multiplies what we had before, so the final result should be 12(choose)4 x 2(choose)1 = 495 x 2 = 990 ways of awarding the prizes. As a side note, the example given in the post above using 4 people happens to work, but it is just a conicidence, there is no mathematically correct reasoning there. You do not divide by 2, you divide by r! where r is the number of identical prizes to be distributed. In the example 2 prizes were chosen, and since 2! = 2 it happened to work. If you look at the expressions for the permute and combine functions, you see that n(choose)r = n! / r!(n-r)! has an extra r! in the denominator when compared to n(permute)r = n! / (n-r)!, this is the r! i just mentioned (which accounts for the difference between unique and idential items being distributed). yep you're right... well done and be careful what you read on these boards! Link to comment Share on other sites More sharing options...
luineul Posted September 21, 2008 Share Posted September 21, 2008 Anyone have any ideas for what I should put on my grad write-up? This is really lame. Link to comment Share on other sites More sharing options...
binkybonks Posted September 21, 2008 Share Posted September 21, 2008 Anyone have any ideas for what I should put on my grad write-up? This is really lame. Link to comment Share on other sites More sharing options...
niicole_ Posted September 22, 2008 Share Posted September 22, 2008 Omg. The subject is NOT GYM! People piss me off when they say that. Its PE. Physical Education. Gym is where you do PE. Thats like saying.. "It'd be funny if someone asked for help in the room" .. Sorry. Just had to state that. Link to comment Share on other sites More sharing options...
marleau_12 Posted September 22, 2008 Share Posted September 22, 2008 ovie8, you're in grade 8 right? Do you have Statz? Link to comment Share on other sites More sharing options...
I♥Wellwood Posted September 22, 2008 Share Posted September 22, 2008 Well, tomorrow I have Socials, English, Chemistry, and French. And then the next day I have Math, Accounting, Psychology, and Biology. Sometimes I wish we were on the semester system.. Link to comment Share on other sites More sharing options...
iheartluongo Posted September 22, 2008 Share Posted September 22, 2008 Aww that sucks. The semester system is much better. I can't even imagine having math and science in the same semester. How did you do on the provincials, those must have been hard the way your school does things. Link to comment Share on other sites More sharing options...
nuckle_191 Posted September 22, 2008 Share Posted September 22, 2008 OK so I am reading Animal Farm by George Orwell, and I need to find 3 examples of connotative language that Major uses in his speech in chapter one to sway the opinions of his listeners thanks EDIT: Heres a link to chapter 1 Thanks Again Link to comment Share on other sites More sharing options...
iheartluongo Posted September 22, 2008 Share Posted September 22, 2008 What are some reasons we do/don't need a Senate? I'm not good w/ government stuff. Any help would be verrrrrry appreciated. =) Link to comment Share on other sites More sharing options...
13Sundin13 Posted September 25, 2008 Share Posted September 25, 2008 *please help* Solve. 7k-(4-2k)=3(k+5)-1 ...the answer is 3, but i don't get it Link to comment Share on other sites More sharing options...
iheartluongo Posted September 25, 2008 Share Posted September 25, 2008 Never mind. The variables are totally messed up. Are you sure that's the question? Link to comment Share on other sites More sharing options...
GoaltenderInterference Posted September 25, 2008 Share Posted September 25, 2008 *please help* Solve. 7k-(4-2k)=3(k+5)-1 ...the answer is 3, but i don't get it Link to comment Share on other sites More sharing options...
iheartluongo Posted September 25, 2008 Share Posted September 25, 2008 OH, I went 3k+5! I didn't multiply 3x5. Silly me. Edit: You see, these are the stupid mistakes I make on my Math tests. Link to comment Share on other sites More sharing options...
WiLBoY Posted September 25, 2008 Share Posted September 25, 2008 *please help* Solve. 7k-(4-2k)=3(k+5)-1 ...the answer is 3, but i don't get it Link to comment Share on other sites More sharing options...
13Sundin13 Posted September 26, 2008 Share Posted September 26, 2008 Thank you very much for your previous help! I have two more problems that will pretty much sum up all that i need to know. If you can explain how to simplify these that would be GREAT!! (-3√3)(-4√5)(2√12) and 3y/y^2+5y-14 + 4/y-2 THANK YOU!! Link to comment Share on other sites More sharing options...
WiLBoY Posted September 26, 2008 Share Posted September 26, 2008 Thank you very much for your previous help! I have two more problems that will pretty much sum up all that i need to know. If you can explain how to simplify these that would be GREAT!! (-3√3)(-4√5)(2√12) and 3y/y^2+5y-14 + 4/y-2 THANK YOU!! Link to comment Share on other sites More sharing options...
Deigler Posted September 26, 2008 Share Posted September 26, 2008 ooh numbers.. Link to comment Share on other sites More sharing options...
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