My Math Forum PERMUTATION

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 12,211 Thanks: 521 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

Posts: 14,413
Thanks: 1024

Re: PERMUTATION

Quote:
 Originally Posted by talo0000 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
Talo, if you think 30/1680 = .0178... is a possibility, then you're definitely not ready fot this,
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: 12,211 Thanks: 521 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: $N={8 \choose 4}=70$ You found the number of permutations, rather than combinations. For a sub-committee 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: $N_1={6 \choose 2}=15$ and so we have: $P(X)=\frac{15}{70}=\frac{3}{14}$ 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: $N_2={5 \choose 4}=5$ $P(X)=\frac{5}{70}=\frac{1}{14}$

 Tags permutation

 Thread Tools Display Modes Linear Mode

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

 Contact - Home - Forums - Cryptocurrency Forum - Top