wildorchidz Posted October 29, 2007 Share Posted October 29, 2007 Hi.... I found these questions sometime back from this site.... I tried doin them n i really went mad.... I never even got an answer nearer to the ones mentioned.... Plz do reply soon... My Gmat is scheduled on 5th Nov.... 1.There are 9 candidates being interviewed for a position. Interviews can only be conducted on the days Monday to Wednesday, two interviews per day. The first interview being conducted each day will have two candidates being interviewed together. How many different schedules are possible for a given period? (A) 9! (B) 9!/2! © 9!/3! (D) 9!/23 (E) 9!/25 2.Lahnie is scheduling her weekend film fest and plans to watch 5 different movies over the weekend. She will watch 2 on Saturday and 3 on Sunday. If she has a library of 6 movies to select from then how many different schedules could she create for her film fest? (A) 7776 (B) 720 © 120 (D) 30 (E) 6 3.Six doctors, six nurses and six anesthesiologists are at a conference. In one of the sessions the group must form teams of two doctors, two nurses and an anesthesiologist. How many such committees can be formed if one of the nurses refuses to work with one of the doctors? (A) 2400 (B) 1800 © 1200 (D) 600 (E) 216 4.At the Chien Dog Show, there were 11 dogs in the semi-finals, of which 4 will be selected for the finals. If two of the dogs selected will be Felix and Shadow, then how many groups of dogs are possible for the finals at the Chien Dog Show? (A) 36 (B) 72 © 720 (D) 3024 (E) 7920 5.A basketball team, which is composed of 3 centers, 6 forwards and 3 guards leaves their seats on the bench for the singing of the national anthem. All but 5 players, two guards, two forwards, and a center, sit down after the anthem, how many different combinations of the players on the bench are there? (A) 120 (B) 792 © 7! (D) 12! / 5! (E) 12! 6.Danielle can choose one or more of the following for her salad: tomatoes, rhubarbs, walnuts. If she selects one or more, how many different combinations are possible? (A) 12 (B) 9 © 7 (D) 6 (E) 5 7.Cody, an avid basketball player, will always wear one or more of the following articles of “lucky” clothing to play basketball: an old sweatband, black tube socks, a “Mugsey” jersey, or his mother’s basketball shorts. If Cody is getting dressed to play basketball, how many different combinations of these articles of clothing are possible? (Assume the order of the clothing does not matter.) (A) 256 (B) 24 © 15 (D) 12 (E) 4 8.Runners V, W, X, Y and Z are competing in the Bayville local triathlon. How many different outcomes of the race are there where X completes the race ahead of Y? (A) 10 (B) 44 © 60 (D) 120 (E) 3125 9.How many different numbers can be created using the digits 7, 3, 4, 2 and 9, if 3 must be kept in the hundreds place and each digit is only used once? (A) 120 (B) 60 © 36 (D) 28 (E) 24 10.A detective has two suspects for a crime and must create a line-up from left to right that consists of two suspects and three decoys (in any order). If all possible line-ups will include both suspects and the decoy spots will be filled from the other 5 officers in the precinct, then how many different line-ups can the detective create? (A) 420 (B) 210 © 120 (D) 35 (E) 3 11.A test prep company is sending 4 members of its offices staff to a conference. The staff consists of 3 directors and 5 assistant directors. If the group sent must consist of 1 director and 2 assistant directors, how many different such groups are possible? (A) 150 (B) 75 © 70 (D) 60 (E) 30 12.A team coach must choose five players from a roster of 12 to start a game. The team must be formed of at least two guards, at least one forward, and at most one center. If team roster has 3 centers, 6 forwards and 3 guards, how many different starting line ups can the team coach form? (A) 36 (B) 180 © 540 (D) 678 (E) 1968 Quote Link to comment Share on other sites More sharing options...
vinuharan Posted October 31, 2007 Share Posted October 31, 2007 can someone please explain the problems? Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q3) The ans is C... 6 D's, 6 N's and 6 A's. First let's form teams of 2 D's, 2 N's and 1 A. This can be done in: 6C2*6C2*6C1 = 1350 ways Now.. 1 particular N refuses to work with 1 particular D. Let's form teams where this particula N and D are included. The particular D can be picked in 1 way. The second D can be picked from the rest 5 D in 5C1 way. Similarly for N. Hence, it is: (1*5C1)*(1*5C1)*(6C1) = 150 Therefore, the ans is: 1350 - 150 = 1200. © Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q4) The ans is A. 11 dogs out of which 4 to be selected. But, Felix and Shadow are already decided. Hence, 2 out of remaining 9 dogs. (1)*(1)*(9C2) = 36 Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q2) Ans seems to be B. This is a Permutation problem as we need to create a schedule. 6P2*4P3= 720. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q1) For this I am getting the answer as A. (guys...pls correct me if I am wrong) TOtal Candidates = 9 Day1:Int1(2 Candidates), Int2(1 Candidate) = 9P2*7P1 Day2:Int1(2 Candidates), Int2(1 Candidate) = 6P2*4P1 Day3:Int1(2 Candidates), Int2(1 Candidate) = 3P2*1P1 Multiplying all the three we get 9!. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q6) Its C... 3C1+3C2+3C3 = 7 Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q7) Its C 4C1+4C2+4C3+4C4 = 15 Quote Link to comment Share on other sites More sharing options...
coyote Posted October 31, 2007 Share Posted October 31, 2007 Q1) For this I am getting the answer as A. (guys...pls correct me if I am wrong) TOtal Candidates = 9 Day1:Int1(2 Candidates), Int2(1 Candidate) = 9P2*7P1 Day2:Int1(2 Candidates), Int2(1 Candidate) = 6P2*4P1 Day3:Int1(2 Candidates), Int2(1 Candidate) = 3P2*1P1 Multiplying all the three we get 9!. :2cents:I was thinking on the same lines but I think we are repeating certain possibilities here. The order of the two candidates in the first slot of any of the 3days does not matter. I think in the 9! part we assume both such possibilities day 1: A B ; C (or) day 1: B A ; C (where A,B, C are candidates) If this is true we need to eliminate those repetitive conditions. Don't know how :doh: Let me know if I'm wrong. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q8) Its C. 5 runners - V,W,X,Y,Z. X has to complete race ahead of Y. Following arrangements possible: 1) X comes first... = 1*4! 2) X comes second = (3P1*1*3!)....(here 1st position can be taken by anyone except X and Y.. 3) X comes third = (3P2*1*2!) 4) X comes fourth and Y comes fifth = (3P3*1.1) X cannot come fifth. Adding the above, ans is 60. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Hi coyote, For 1) as the ques. talks about interview schedules and hence order of candidates should matter. I could'nt think of any other way....:( Quote Link to comment Share on other sites More sharing options...
coyote Posted October 31, 2007 Share Posted October 31, 2007 Of course Ketan, the order matters. But two candidates when interviewed "together" in the same slot, then the order should not matter. I am referring to the first slots on each day. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q9) The ans is B provided we assume that 3 digit, 4 digit and 5 digit numbers can be created. 3 digit numbers = (1)*(4P2) = 12 4 digit numbers = (4P1)*1*(3P2) = 24 5 digit numbers = (4P2)*1*(2P2) = 24 Adding above = 60 Quote Link to comment Share on other sites More sharing options...
vinuharan Posted October 31, 2007 Share Posted October 31, 2007 :2cents:I was thinking on the same lines but I think we are repeating certain possibilities here. The order of the two candidates in the first slot of any of the 3days does not matter. I think in the 9! part we assume both such possibilities day 1: A B ; C(or) day 1: B A ; C (where A,B, C are candidates) If this is true we need to eliminate those repetitive conditions. Don't know how :doh: Let me know if I'm wrong. since we are finding the possible number of schedules,i think we should treat AB and BA as different. If we want to remove the repetitions we should go for combination.Even then i dont get E as answer. Quote Link to comment Share on other sites More sharing options...
coyote Posted October 31, 2007 Share Posted October 31, 2007 They have said the 'candidates will be interviewed "together" ' .I think we should consider them as a pair and not individually.:confused: I,too, don't see how to put this down mathematically. Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 Q11) Ans is D 1 D and 3 ADs = 3C1*5C3 2 Ds amd 2 ADs = 3C2*5C2 Adding = 60 Quote Link to comment Share on other sites More sharing options...
ketank Posted October 31, 2007 Share Posted October 31, 2007 For Q 12) My ans is coming 1548. However, it's not matching with the supplied options. Could'nt understand Q10) properly. Quote Link to comment Share on other sites More sharing options...
vinuharan Posted October 31, 2007 Share Posted October 31, 2007 10) 7*6*5*2*1=420 For the first three positions i considered both police officers and criminals and for the last two position i considered only 2 criminals. correct me if iam wrong... Quote Link to comment Share on other sites More sharing options...
wildorchidz Posted November 1, 2007 Author Share Posted November 1, 2007 Thanks guys for all the explanation .. I was really goin mad with these... Quote Link to comment Share on other sites More sharing options...
ketank Posted November 2, 2007 Share Posted November 2, 2007 welcome :) Quote Link to comment Share on other sites More sharing options...
DWarrior Posted November 2, 2007 Share Posted November 2, 2007 They have said the 'candidates will be interviewed "together" ' .I think we should consider them as a pair and not individually.:confused: I,too, don't see how to put this down mathematically. I also did it your way and assumed two people being interviewed together are picked without regard to order. My result was: Day 1: 9C2*7 Day 2: 6C2*4 Day 3: 3C2*1 Thus: 9!/2^3 If you also allow second interview to be conducted either individually or in pairs, that still doesn't give any of the answers listed. Quote Link to comment Share on other sites More sharing options...
coyote Posted November 2, 2007 Share Posted November 2, 2007 If you also allow second interview to be conducted either individually or in pairs, that still doesn't give any of the answers listed. Precisely. I tried several ways. But none yields 25 in the denominator whatsoever. Can we assume that the answer is wrong? ;) Quote Link to comment Share on other sites More sharing options...
Chlyxu Posted April 5, 2012 Share Posted April 5, 2012 Do we have a verdict on what the final correct answer for Q1 might be? I haven't been able to get a satisfactory answer myself > Quote Link to comment Share on other sites More sharing options...
Tristar Posted May 24, 2015 Share Posted May 24, 2015 (edited) Q1.There are 9 candidates being interviewed for a position. Interviews can only be conducted on the days Monday to Wednesday, two interviews per day. The first interview being conducted each day will have two candidates being interviewed together. How many different schedules are possible for a given period? Ans: We have three interviews in all: one on Monday, one on Tuesday and one on Wednesday. We are just considering the first interview to be held on each of these three days. For Monday, there will be 9C2 = 36 different schedules involving BOTH candidates. For, Tuesday, there will be 7C2 = 21 such schedules and for Wednesday 5C2 = 10 different schedules. All in all, there will be 36*21*10= 7,560 different schedules. Edited May 24, 2015 by Tristar Quote Link to comment Share on other sites More sharing options...
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