# Thread: 12 Permutation n Combinations problems

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## 12 Permutation n Combinations problems

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!
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 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
(C) 540
(D) 678
(E) 1968

2. Good post? |
can someone please explain the problems?

3. Good post? |
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. (C)

4. Good post? |
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

5. Good post? |
Q2) Ans seems to be B.
This is a Permutation problem as we need to create a schedule.
6P2*4P3= 720.

6. Good post? |
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!.

7. Good post? |
Q6) Its C...
3C1+3C2+3C3 = 7

8. Good post? |
Q7) Its C
4C1+4C2+4C3+4C4 = 15

9. Good post? |
Originally Posted by ketank
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!.
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

Let me know if I'm wrong.

10. Good post? |
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.