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 November 21st, 2011, 05:48 PM #1 Newbie   Joined: Nov 2011 Posts: 7 Thanks: 0 Permutation and combination ? A professor has a class of 7 graduates, consists of 4 men and 3 women. (I). Professor wishes to line up all the graduates from left to right on a bench for the class photo.In how many ways this be done? (ii) how many line ups are possible if no two mens may sit together and no two women may sit together? (iii) two of the students, Rio and James, cannot stop bugging each other. How many line-ups are there in which rio and james are not sitting beside each other? (iv) The professor required to choose two graduates,one male and one female, to be class representatives.In how many ways may this be done? (v) The professor must select 3 of the students to help her with a certain project.how many possible selections are there? Can any body solve this
 November 21st, 2011, 06:25 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Permutation and combination ? (i) Using the fundamental counting principle, we find there are 7 choices for the first seat, 6 for the 2nd, 5 for the third and so on. Thus, the number of ways is: $7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=7!=5040$ (ii) Letting M represent male and F represent female, we must have: M F M F M F M This means the number of ways will be given by: $4!\cdot3!=144$ (iii) First putting James on the left of Rio, there are 5 ways for there to be 1 person between them and 5! different lineups of the other people. Including the lineups with Rio on the left of James this is a total of: $2\cdot5\cdot5!$ With two people between them, we find: $2\cdot4\cdot5!$ With 3 in between: $2\cdot3\cdot5!$ With 4 in between: $2\cdot2\cdot5!$ With 5 in between: $2\cdot1\cdot5!$ Adding, we find: $2\cdot5!$$5+4+3+2+1$$=30\cdot5!=3600$ Thus, there are 3600 ways James and Rio are not seated together. (iv) The professor has 4 males to choose from and 3 females, thus there are: $4\cdot3=12$ There are 12 ways to choose 1 male and 1 female. (v) Here we use the combination function: ${7 \choose 3}=35$ Thus, we find there are 35 ways to choose 3 people from a group of 7.
November 21st, 2011, 06:42 PM   #3
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Re: Permutation and combination ?

Quote:
 Originally Posted by MarkFL (iii) First putting James on the left of Rio, there are 5 ways for there to be 1 person between them and 5! different lineups of the other people. Including the lineups with Rio on the left of James this is a total of: $2\cdot5\cdot5!$ With two people between them, we find: $2\cdot4\cdot5!$ With 3 in between: $2\cdot3\cdot5!$ With 4 in between: $2\cdot2\cdot5!$ With 5 in between: $2\cdot1\cdot5!$ Adding, we find: $2\cdot5!$$4+3+2+1$$=20\cdot5!=2400$ Thus, there are 2400 ways James and Rio are not seated together.
Mark,

I'm going to disagree with your answer. I get 3600 in two different ways. Call the people A,B,C,D,E,J and R.

Solution 1: There are 7! = 5040 ways to line up everyone.
There are 2 ways to line up James and Rio side-by-side. There are now six ways to line up these "objects" A,B,C,D,E,(JR) or (RJ). Thus there are 2*6! = 1440 ways to line up James and Rio together.
Thus there are 5040 - 1440 = 3600 ways to line them up apart.

Solution 2: Line up everyone except James and Rio. There are 5! = 120 ways to do that. Call these five people A,B,C,D,E.
Space them out _A_B_C_D_E_. There are six potential spaces for James and Rio to go into. You can insert them in 6*5 = 30 ways.
Thus both tasks can be done in 120*30 = 3600 ways.

 November 21st, 2011, 07:23 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Permutation and combination ? I have corrected my omission of the first case I considered. Now my result agrees with yours!

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