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June 7th, 2012, 07:06 PM   #1
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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
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June 7th, 2012, 07:11 PM   #2
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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?
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June 8th, 2012, 04:50 AM   #3
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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
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June 8th, 2012, 08:40 AM   #4
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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...
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June 8th, 2012, 10:12 AM   #5
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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 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:

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:



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