securitygeek Posted May 10, 2005 Share Posted May 10, 2005 A lot of people asked me what kind of probability problems I got on the actual exam. Since I signed a non disclosure agreement, I can't reveal the actual questions, but following are two permutation/combinations- probability problems similar to the ones I got on the real GMAT. 1. There is a set of characters, A B C D E F G H I J K. There are 4 character and 3 character codes to be made out of these characters. What is the ratio of the total number possibilities of the 4 character codes to the total possibilities of the 3 character codes? 2. There are 6 red balls and 3 blue balls in a bag. 2 red balls are taken out. What is the possibility that the next ball taken out will be a red ball? ( Yeah, they are that simple) Quote Link to comment Share on other sites More sharing options...
prep2max Posted May 10, 2005 Share Posted May 10, 2005 1. 11c4*41 : 11c3*3! 2. 4/7 Quote Link to comment Share on other sites More sharing options...
haneen97 Posted May 10, 2005 Share Posted May 10, 2005 Is it not the first one is a permutation? I think order matter since ABC is a different code that CBA. I got 8:1 Quote Link to comment Share on other sites More sharing options...
securitygeek Posted May 10, 2005 Author Share Posted May 10, 2005 First one is a permutation problem. I thought it was a combination, but none of the answer choices matched my answer ( thank god). Then I did this the permutations way and my answer matched the answer choice ;) Quote Link to comment Share on other sites More sharing options...
rd_eastbay Posted May 10, 2005 Share Posted May 10, 2005 Thanks Security. Yes (1) is a permutation problem. The answer is 11P4 / 11P3 = 8. Quote Link to comment Share on other sites More sharing options...
jatinet Posted May 10, 2005 Share Posted May 10, 2005 rd_eastbay, 11P4 I know how to solve combinations but not sure of permutations will appreciate if you could explain a bit about solving such permutaions Quote Link to comment Share on other sites More sharing options...
rd_eastbay Posted May 11, 2005 Share Posted May 11, 2005 Sure. First the formula : nPk = n!/(n-k)!. I assume you already knew that but it bears repeating. So 11P4 = 11!/7! = 11 x 10 x 9 x 8 And one has to use permutations instead of combinations because (as explained earlier), even if the same alphabets are picked (say A, B, C) they can be arranged in different ways. Ie whenever "order" matters, use permutations. HTH. rd_eastbay, 11P4 I know how to solve combinations but not sure of permutations will appreciate if you could explain a bit about solving such permutaions Quote Link to comment Share on other sites More sharing options...
jatinet Posted May 11, 2005 Share Posted May 11, 2005 Thanks rd Quote Link to comment Share on other sites More sharing options...
lego2401 Posted May 11, 2005 Share Posted May 11, 2005 1- 11*10*9*8/11*10*9 = 8 2- 4/7 Yes, from my experience, this is pretty much the standard of prob/comb/perm Qs on the GMAT. No matter which range your in, it wont get more difficult than this. Don't work too much on these questions, you'll waste valuable time that can be used to enhance other skills for the GMAT. This is pretty much as hard as it gets. Quote Link to comment Share on other sites More sharing options...
rd_eastbay Posted May 11, 2005 Share Posted May 11, 2005 Lego / Security, Any examples of the algebra/number-thoery problems. While conceptually not hard, I find these real difficult to get right under time pressure. I am sure many other people feel the same way... 1- 11*10*9*8/11*10*9 = 8 2- 4/7 Yes, from my experience, this is pretty much the standard of prob/comb/perm Qs on the GMAT. No matter which range your in, it wont get more difficult than this. Don't work too much on these questions, you'll waste valuable time that can be used to enhance other skills for the GMAT. This is pretty much as hard as it gets. Quote Link to comment Share on other sites More sharing options...
prep2max Posted May 11, 2005 Share Posted May 11, 2005 11c4*4! : 11c3*3! = 8 : 1 Quote Link to comment Share on other sites More sharing options...
arjmen Posted May 11, 2005 Share Posted May 11, 2005 1. 11p4/11p3 = (11!/7!)*(8!/11!) = 8!/7! = 8 2. (6-2)/(9-2) = 4/7 Quote Link to comment Share on other sites More sharing options...
akshayxyz Posted October 19, 2005 Share Posted October 19, 2005 1. There is a set of characters, A B C D E F G H I J K. There are 4 character and 3 character codes to be made out of these characters. What is the ratio of the total number possibilities of the 4 character codes to the total possibilities of the 3 character codes? Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated) similarly # of 3 character codes = 11^3 so the ratio is 11:1 b) another approach, once again considering that duplicate characters are allowed.. # of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes).. we have 11 choices so.. its again 11:1 though I agree that if duplicates are not allowed, the ratio has to be 8:1 Hope I am not wrong. I am taking the test in 2 days :), that said, any magic potions for Verbal ;) Quote Link to comment Share on other sites More sharing options...
blueskies Posted November 8, 2005 Share Posted November 8, 2005 1. 11c4*41 : 11c3*3! 2. 4/7 Why are you multiplying the combinations by 4! and 3! respectively? I believe the solution is simply 11C4 / 11C3. Ashayxyz, I see your point, but the feeling I get from the question is that letters are not to be duplicated. Of course, if "8:1" and "11:1" both show up in the answer choices... we're screwed. :) Quote Link to comment Share on other sites More sharing options...
gonags Posted November 19, 2005 Share Posted November 19, 2005 Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated) similarly # of 3 character codes = 11^3 so the ratio is 11:1 b) another approach, once again considering that duplicate characters are allowed.. # of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes).. we have 11 choices so.. its again 11:1 though I agree that if duplicates are not allowed, the ratio has to be 8:1 Hope I am not wrong. I am taking the test in 2 days :), that said, any magic potions for Verbal ;) -------------- Securitygeek, what do you think about Akshayxyz's post? I think he's got a valid point. Cheers! Quote Link to comment Share on other sites More sharing options...
CTG1983 Posted November 19, 2005 Share Posted November 19, 2005 I also think akshay is right. Example: How many telephone connections can be made with 5 digits from the natural numbers 1 to 9 inclusive? The ans. is 9^5. Quote Link to comment Share on other sites More sharing options...
mav Posted November 27, 2005 Share Posted November 27, 2005 Yeah, I agree that the proportion should be 11^4 : 11^3 = 11 At first, I think it's 11P4 : 11P3 = 8, but I think this approach is wrong. Quote Link to comment Share on other sites More sharing options...
preity Posted November 27, 2005 Share Posted November 27, 2005 I think ETS will not give 11 in one of the answer choices..As the problem is not that clear about repetetions part. (Afterall its a problem solving Question. So go with the answers) If they give 11 and 8 as answer choices then it will raise a conflict..I think they either give 8 or 11 as answer choices(not both) What do u say securitygeek????? Quote Link to comment Share on other sites More sharing options...
CTG1983 Posted November 27, 2005 Share Posted November 27, 2005 Preity, Why do u think so? Quote Link to comment Share on other sites More sharing options...
deren Posted November 27, 2005 Share Posted November 27, 2005 Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated) similarly # of 3 character codes = 11^3 so the ratio is 11:1 b) another approach, once again considering that duplicate characters are allowed.. # of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes).. we have 11 choices so.. its again 11:1 though I agree that if duplicates are not allowed, the ratio has to be 8:1 Hope I am not wrong. I am taking the test in 2 days :), that said, any magic potions for Verbal ;) hey akshayxyz; first of all your solution is wrong because when you do this in your way ; 11*11*11*11 => solution set includes 4 AAAA and 4 BBBB and so on. Understand. So if we solve the problem in your way we need to substract 11 X 3 from our solution set. and the answer becomes something like that: (11^4-33)\(11^3-22) .... and the solution is not even a integer... I think this is a perm question and the real answer must be 8:1 ... we don't know about the choices but probably your solution is Quote Link to comment Share on other sites More sharing options...
nickel Posted November 28, 2005 Share Posted November 28, 2005 Guys, Akshay is correct. We use permutation and combination formulas when choosing the 2nd number is dependent on the 1st number. The answer should be 11. Let me give you a simple example. How many 4 digit key codes can we make using numbers from 0-9. We all know that 0-9999 is 10,000 numbers right ? 10*10*10*10 = 10,000. Now doing 10P5 = 5040 codes. I hope this clears up the confusion Quote Link to comment Share on other sites More sharing options...
ChhotiAankhe_BadeSapane Posted December 9, 2005 Share Posted December 9, 2005 When I started reading this post I was so surprised to how so many ppl are finalizing answer 8. It has to be 11. 4 Char codes I can make in ..11 * 11 * 11 * 11 3 Char codes I can make in ..11 * 11 * 11 Ratio is 11. One suggestion to all... don't always fall in trap of formulaes to PROB Qs. You may apply common sense and you will be through! Good Luck! Quote Link to comment Share on other sites More sharing options...
deren Posted December 9, 2005 Share Posted December 9, 2005 When I started reading this post I was so surprised to how so many ppl are finalizing answer 8. It has to be 11. 4 Char codes I can make in ..11 * 11 * 11 * 11 3 Char codes I can make in ..11 * 11 * 11 Ratio is 11. One suggestion to all... don't always fall in trap of formulaes to PROB Qs. You may apply common sense and you will be through! Good Luck! so you are the new math professor ha :crazy::boxing: Quote Link to comment Share on other sites More sharing options...
heroesvillains Posted September 5, 2007 Share Posted September 5, 2007 I know this is an old thread but it has really cleared up some things for me so thanks! Quote Link to comment Share on other sites More sharing options...
Gnid Posted September 5, 2007 Share Posted September 5, 2007 i dont know how to get answer 2...can someone explain why answer two is (6-2)/(9-2) Quote Link to comment Share on other sites More sharing options...
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