June 7th, 2012, 07:06 PM  #1 
Newbie Joined: May 2012 Posts: 9 Thanks: 0  PERMUTATION
committee with eight members A,B,C,D,E,F,G, AND H DECIDE TO APPOINT SUBCOMMITTEE OF FOUR MEMBERS WHAT IS THE PROBABILITY THAT 1  A AND B ARE IN THE COMMITTEE 2 A, B AND C ARE NOT IN THE COMMITTE 
June 7th, 2012, 07:11 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: PERMUTATION
How may subcommittees in total are possible? Hint: We are choosing 4 from 8... 1.) How many will have A and B in them? Hint: How many are left to choose and how many are we choosing? 2.) How many will not have A, B and C in them? Hint: How many are left to choose and how many are we choosing? 
June 8th, 2012, 04:50 AM  #3 
Newbie Joined: May 2012 Posts: 9 Thanks: 0  Re: PERMUTATION
thank you do you mean a ) n(s)=p(8,4)=1680 n(E)=p(2,2). p(6,2)=30 P(E)= 30/1680 b) n(E1)=p(5,4)=120 p(E1)=120/1680 
June 8th, 2012, 08:40 AM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,148 Thanks: 685  Re: PERMUTATION Quote:
and need help from your teacher...  
June 8th, 2012, 10:12 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: PERMUTATION
How may subcommittees in total are possible? Hint: We are choosing 4 from 8... The number of ways N to choose 4 from 8 is given by: You found the number of permutations, rather than combinations. For a subcommittee the order in which the people are chosen doesn't matter, you will still have the same 4 people. 1.) How many will have A and B in them? Hint: How many are left to choose and how many are we choosing? Since we are given that A and B have been chosen, we have 6 left to choose from and we are choosing 2 from this 6 to fill out the remaining 2 positions. Thus: and so we have: 2.) How many will not have A, B and C in them? Hint: How many are left to choose and how many are we choosing? Here, we only have 5 that we are allowed to choose from, and we are choosing 4, hence: 

Tags 
permutation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Permutation  abramo_k  Algebra  4  February 27th, 2013 12:11 PM 
permutation  kingkos  Algebra  1  December 6th, 2012 12:39 PM 
permutation  panky  Algebra  8  November 21st, 2011 11:11 PM 
Permutation  cyt_91  Advanced Statistics  1  January 29th, 2010 09:12 AM 
Permutation  cyt_91  Advanced Statistics  1  January 29th, 2010 07:16 AM 