Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist, how

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

I am right to think of it like this. t he idea behind adding the overlap is shown in the addition rule:

P(A\/B\/C)=P(A)+P(B)+P(C)-P(A/\B)-P(A/\C)-P(B/\C)+P(A/\B/\C)


My question is. how do you know what P(A/\B/\C) is in my example, based on the above idea (i know am not dealing with probability, but logic is same)
 
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PAllen

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

is their a name for this method. what is the general logic behind it. is it related to complements

e.g

Sample space - (those you dont want) = those you do want

those you dont want are usually a smaller group and easy to work out using this method.


but i wanna know how you knew you have to add 7 and 1 to "those you dont want"

i just cant see. what does 7 and 1 represent. 7 physicists???
I don't know a name for this method. I've just seen it being used in books to derive results, and put it in mind as a useful general approach.

It is possible to have a committee missing both physicists and chemists; and also missing both physicists and biologists. How many of each type? Of the groups you subtracted, how many would include each these doubly missing committees? Is it ok to subract the same committee twice? You should be able to see the justification for the final additions from this.
 
Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

I don't know a name for this method. I've just seen it being used in books to derive results, and put it in mind as a useful general approach.

It is possible to have a committee missing both physicists and chemists; and also missing both physicists and biologists. How many of each type? Of the groups you subtracted, how many would include each these doubly missing committees? Is it ok to subract the same committee twice? You should be able to see the justification for the final additions from this.
ok. i know how to work this out and set it out methamtically. tell me if this is correct.

let: a = phys, b=bio, c=chemist
1. We want "at least 1 of each from 3 groups", let this = E
2. Another way of writing this is 1- [tex]\neg[/tex] E, which is "NOT at least 1 of each from 3 groups)
3. All possibilities: a +b +c+(a/\b) +(b/\c)+(a/\c)

a=(4c6)=0
b=(7c6)=7
c=(6c6)=1

a/\b=(11c6)=462 (this repeats what we did in a, which is,a=a/\not b/\not c)
b/\c=(10c6)=210 (this repeats what we did in b, which is,b=not a/\ b/\not c)
a/\c=(13c6)=1716 (this repeats what we did in c, which is,c=not a/\not b/\c)

TOTAL = 0+7+1+462+210+1716=2396

ENTIRE SAMPLE SPACE = (17c6)=12376

sample space - E = (12376-2369)=9988 (we have already had repeats)

so 9988+0+1+7=9996

What have i doen wrong above, how come i need to add 8 back at end (i have took away 8 instead from start)
 
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PAllen

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

Start from here as you wrote:

17C6=Entire sample space =12376
10C6=From (0 biologists, 4 physicists and 6 chemists) take 6=210
11C6=From (7 biologists, 4 physicists and 0 chemists) take 6=462
13C6=From (7 biologists, 0 physicists and 6 chemists) take 6=1716

12376-210-462-1716=9988

Answer the questions I posed:

How many committees are there no chemists and no physicists? In the terms subtracted above, how many would include such committees? If more than one, you've subtracted the same committees more than once. Compensate.

Same for committees with no biologists and no physicists.

This should lead you directly to the +1 +7.

If there is some part of these questions you don't understand, tell me what exactly is unclear.
 

PAllen

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

ok. i know how to work this out and set it out methamtically. tell me if this is correct.

let: a = phys, b=bio, c=chemist
1. We want "at least 1 of each from 3 groups", let this = E
2. Another way of writing this is 1-E, which is "NOT at least 1 of each from 3 groups)
3. All possibilities: a +b +c+(a/\b) +(b/\c)+(a/\c)

a=(4c6)=0
b=(7c6)=7
c=(6c6)=1

a/\b=(11c6)=462 (this includes missing c)
b/\c=(10c6)=210 (this includes missing a)
a/\c=(13c6)=1716 (this includes missing b)

TOTAL = 0+7+1+462+210+1716=2396

ENTIRE SAMPLE SPACE = (17c6)=12376

sample space - E = (12376-2369)=9988 (we have already had repeats)

so 9988+0+1+7=9996

What have i doen wrong above, how come i need to add 8 back at end (i have took away 8 instead from start)
Your mistake is as follows:

missing c includes missing a and c
missing a includes missing a and c

Thus, missing a and c committees have been included twice. So you must subtract your b (not add it) . Similar for missing a and b.
 
Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

Your mistake is as follows:

missing c includes missing a and c
missing a includes missing a and c

Thus, missing a and c committees have been included twice. So you must subtract your b (not add it) . Similar for missing a and b.
thanks for all your help. now i understand that:

if we add all of the below. it means, we add, a + b + c again (so repeat)

if you have time. can you tell me, what we have repeated in each of these

1.a/\b=(11c6)..........i think we are repeating what we did in "a" and "b"
2.b/\c=(10c6)..........i think we are repeating what we did in "b" and "c"
3.a/\c=(13c6)..........i think we are repeating what we did in "a" and "c"

Also, The part which is unclear to me, is the OVERLAP part, i just cant identify any overlaps (i cant see why, the above combinations has got something we repeated already), please help if you have time
 
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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

Your mistake is as follows:

missing c includes missing a and c
missing a includes missing a and c

Thus, missing a and c committees have been included twice. So you must subtract your b (not add it) . Similar for missing a and b.
b = biologist
c= chemist

are you saying that because a/\b, b/\c, a/\c include a,b,c (so we must subtract the values of a,b and c i.e subtract 0 + 1 + 7)
a=(4c6)(13c0)=0
b=(7c6)(10c0)=7 (7 bio and 0 from the group of phys and chem)
c=(6c6)(11c0)=1 (6 chem and 0 from the group of phys and bio)
 
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PAllen

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

Maybe phrasing in terms of overlap, as you request, will be clearer.

The set of committees with only physicists and biologists includes committees with only biologists.

The set of committees with only chemists and biologists includes committees with only biologists.

To get the count of their union, you must remove the count of their overlap (committees with only biologists).
 
Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

Maybe phrasing in terms of overlap, as you request, will be clearer.

The set of committees with only physicists and biologists includes committees with only biologists.
so a/\b includes the set of 7 biologist

The set of committees with only chemists and biologists includes committees with only biologists.
b/\c, includes the set of 7 biologist AGAIN.


so we must get rid of 7 biologists

LIKEWISE a/\b,.......include 4 physiists and and a/\c 4 physicicsts again !


so why dont we subtract 7+4....why is it 7+1?

or am i thinking of it wrong
 

PAllen

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Re: Permuations Questions. Science comitte has 7 bioligist, 4 physcicist, 6 chemist,

so a/\b includes the set of 7 biologist


b/\c, includes the set of 7 biologist AGAIN.


so we must get rid of 7 biologists

LIKEWISE a/\b,.......include 4 physiists and and a/\c 4 physicicsts again !


so why dont we subtract 7+4....why is it 7+1?

or am i thinking of it wrong
You are thinking of it wrong. You are not getting rid of 7 biologists (that's mean). You are getting rid of the 7 possible committees of only biologists (never hurts to get rid of a committee). There are no committees of only physicists, as you have noted in an earlier post. However, there is one committee of only chemists, that is also doubly counted. Thus, 7+1 doubly counted committees.
 

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