Possible Science Committees from 17 Scientists

[cloud360]

Homework Statement

A science committee of 6 people is to be formed from a group
of 17 scientists (7 biologists, 4 physicists and 6 chemists). How
many possible science committees are there:-


Q.Given that the committee must consist of at least 1 biologist,
1 physicist and 1 chemist

Homework Equations

none

The Attempt at a Solution

I first found the total sample space , which is (17C6) = 12376.

I then found the probability (edit, sorry meant permutation) of 0 bioligists, which is (7C0)*(10C6) = 210

Entire sample space - permuation of 0 biologist = 12166 (this should give at least 1 biologist)

I did this for the rest, i.e 0 chemist, 0 physicists ...

I dotn even know if am doing the right thing. Can somone tell me how i work this out

4. Completed Solutions

4 Physicists
4 phys + 1 bio + 1 chem= (4C4)*(7C1)*(6C1) = 42


3 Physicists
1st possibility = 3 phys + 2 bio + 1 chem = (4C3)*(7C2)*(6C1) = 504
2nd possibility = 3 phys + 1 bio + 2 chem = (4C3)*(7C1)*(6C2) = 420

504 + 420 = 924

2 Physicists
1st possibility = 2 phys + 2 bio + 2 chem = (4C2)*(7C2)*(6C2) = 1890
2st possibility = 2 phys + 1 bio + 3 chem = (4C2)*(7C1)*(6C3) = 840
3rd possibility = 2 phys + 3 bio + 1 chem = (4C2)*(7C3)*(6C1) = 1260

1890 +840 +1260 = 3990

1 Physicists
1st possibility = 1 phys + 4 bio + 1 chem = (4C1)*(7C4)*(6C1) = 840
2nd possibility = 1 phys + 3 bio + 2 chem = (4C1)*(7C3)*(6C2) = 2100
3rd possibility = 1 phys + 2 bio + 3 chem = (4C1)*(7C2)*(6C3) = 1680
4th possibility = 1 phys + 1 bio + 4 chem = (4C1)*(7C1)*(6C4) = 420

840 +2100 +1680 +420 = 5041

Total

42 + 924 + 3990 + 5041 = 9997

 

[tiny-tim]

welcome to pf!

hi cloud360! welcome to pf!

A science committee of 6 people is to be formed from a group
of 17 scientists (7 biologists, 4 physicists and 6 chemists).

How many possible science committees are there:-
Q.Given that the committee must consist of at least 1 biologist,
1 physicist and 1 chemist

(forget probabilities … the question doesn't ask for them! )

since the number of physicists is the smallest, i'd start with them …

eg if there's all 4 physicists, then there's exactly one chemist and one biologist

 

[cloud360]

hi cloud360! welcome to pf!


(forget probabilities … the question doesn't ask for them! )

since the number of physicists is the smallest, i'd start with them …

eg if there's all 4 physicists, then there's exactly one chemist and one biologist

Following what you said, would mean:

4 physicists = (4C4)=1

1 Biologist = (7C1) = 1

1 Chemist = (6C1) =1

If we multiple together, we get 1 * 1 * 1= 1

Does this mean the answer is 1? if so, can you explain why, because i just can't see.

(ALSO, I AM SO HAPPY YOU REPLIED, THANKS FOR YOUR TIME)

 

[tiny-tim]

hi cloud360!

that's right … the number of committees with 4 physicists is 1

but now you need the number with 3, 2, and 1 physicist (which isn't so easy! ) …

have a go! ​

 

[cloud360]

hi cloud360!

that's right … the number of committees with 4 physicists is 1

but now you need the number with 3, 2, and 1 physicist (which isn't so easy! ) …

have a go! ​

4 Physicists
Why is the answer to 4 physicists + 1 bioligist + 1 chemist = 1

isnt it supposed to be:

4 physicists (4C4)=1
2 others (13C2) = 78

1*78 = 78



(i chose 13 because, 13 is the ones which are not physicists. How come this is not correct?)

 

[cloud360]

Anyway, carry on with you method

4 Physicists
4 phys + 1 bio + 1 chem= (4C4)*(7C1)*(6C1) = 42


3 Physicists
1st possibility = 3 phys + 2 bio + 1 chem = (4C3)*(7C2)*(6C1) = 504
2nd possibility = 3 phys + 1 bio + 2 chem = (4C3)*(7C1)*(6C2) = 420

504 + 420 = 924

2 Physicists
1st possibility = 2 phys + 2 bio + 2 chem = (4C2)*(7C2)*(6C2) = 1890
2st possibility = 2 phys + 1 bio + 3 chem = (4C2)*(7C1)*(6C3) = 840
3rd possibility = 2 phys + 3 bio + 1 chem = (4C2)*(7C3)*(6C1) = 1260

1890 +840 +1260 = 3990

1 Physicists
1st possibility = 1 phys + 4 bio + 1 chem = (4C1)*(7C4)*(6C1) = 840
2nd possibility = 1 phys + 3 bio + 2 chem = (4C1)*(7C3)*(6C2) = 2100
3rd possibility = 1 phys + 2 bio + 3 chem = (4C1)*(7C2)*(6C3) = 1680
4th possibility = 1 phys + 1 bio + 4 chem = (4C1)*(7C1)*(6C4) = 420

840 +2100 +1680 +420 = 5041

Total

42 + 924 + 3990 + 5041 = 9997

 

[tiny-tim]

Why is the answer to 4 physicists + 1 bioligist + 1 chemist = 1

isnt it supposed to be:

4 physicists (4C4)=1
2 others (13C2) = 78

1*78 = 78



(i chose 13 because, 13 is the ones which are not physicists. How come this is not correct?)

oops! i forgot what the question was

i was copying your …

1 Biologist = (7C1) = 1

1 Chemist = (6C1) =1

… which should have been

1 Biologist = (7C1) = 7

1 Chemist = (6C1) = 6,

giving a total of 42

(your 13C2 is wrong because it includes committees with 2 of one and none of the other )

 

[tiny-tim]

1st possibility = 3 phys + 2 bio + 1 chem = (4C3)*(7C2)*(6C1) = 504
2nd possibility = 3 phys + 1 bio + 2 chem = (4C3)*(7C1)*(6C2) = 420

504 + 420 = 924

Yup!

 

[cloud360]

Yup!

ok, i got

9997

i must have taken up enough of your time already. so please, if you have time.next time you come online, can you kindly tell answer these:

1st question:Is my method above correct


2nd question:Is their a faster way to do it (thats an awful lot of working i did, just give me names of certain methods and i will go learn myself)


3rd question. The event with 4 phys has 1 possibilites, The event with 3 phys has 2 possibilities, The event with 2 phys has 3 possibilites, The event with 1 phys has 4 possibilites, IS THEIR A METHOD TO KNOW EXACTLY HOW MANY POSSIBILTIES I SHOULD EXPECT FOR EACH


I am really grateful for all the help you provided, thanks again for the guidance

 

[PAllen]

I think 9997 is almost (but not quite) right. I would suggest another approach.

Total committess of 6, any composition.

Take away committees missing all physicists, missing all biologists, missing all chemists.

But these subtractions doubly remove committees missing both physicists and chemists; and also missing both physicists and biologists. Compensate for this.

 

[cloud360]

I think 9997 is almost (but not quite) right. I would suggest another approach.


Take away committees missing all physicists, missing all biologists, missing all chemists.

isnt this is impossible. if we don't choose a phys or a chem or a bio, what can we chooose then? we have to choose 1 of them. how can we choose 0 of each?


Do you know answer to this? The event with 4 phys has 1 possibilites, The event with 3 phys has 2 possibilities, The event with 2 phys has 3 possibilites, The event with 1 phys has 4 possibilites (see post 6), IS THEIR A METHOD TO KNOW EXACTLY HOW MANY POSSIBILTIES I SHOULD EXPECT FOR EACH

 

[PAllen]

isnt this is impossible. if we don't choose a phys or a chem or a bio, what can we chooose then? we have to choose 1 of them. how can we choose 0 of each?


Do you know answer to this? The event with 4 phys has 1 possibilites, The event with 3 phys has 2 possibilities, The event with 2 phys has 3 possibilites, The event with 1 phys has 4 possibilites (see post 6), IS THEIR A METHOD TO KNOW EXACTLY HOW MANY POSSIBILTIES I SHOULD EXPECT FOR EACH

'missing all physicists' is one possibility (e.g. containing only chemists and biologists); missing all biologists means containing only chemists and physicists; I hoped that would be clear.

Note, separately, it is possible to deduce that the final answer is an even number, so 9997 cannot be correct.

 

[PAllen]

Do you know answer to this? The event with 4 phys has 1 possibilites, The event with 3 phys has 2 possibilities, The event with 2 phys has 3 possibilites, The event with 1 phys has 4 possibilites (see post 6), IS THEIR A METHOD TO KNOW EXACTLY HOW MANY POSSIBILTIES I SHOULD EXPECT FOR EACH

In your post #6, you have the right components, but make a silly addition mistake.

However, I am trying to propose a way with *many* fewer individual computations, less room for error, that is also the way one systematically solves much more complex examples of this type. You start with a simple computation that includes cases you don't want. You subract cases you don't want. Most often, this will double subract some cases, so add them back. etc. You get a start, subractions, additions (enough for this case), subractions, additions, etc. Generally, with enormously fewer terms than your post 6 you can compute solutions to complex problems of this type.

 

[PAllen]

isnt this is impossible. if we don't choose a phys or a chem or a bio, what can we chooose then? we have to choose 1 of them. how can we choose 0 of each?


Do you know answer to this? The event with 4 phys has 1 possibilites, The event with 3 phys has 2 possibilities, The event with 2 phys has 3 possibilites, The event with 1 phys has 4 possibilites (see post 6), IS THEIR A METHOD TO KNOW EXACTLY HOW MANY POSSIBILTIES I SHOULD EXPECT FOR EACH

In your post #6, you have the right components, but make a silly addition mistake.

However, I am trying to propose a way with *many* fewer individual computations, less room for error, that is also the way one systematically solves much more complex examples of this type. You start with a simple computation that includes cases you don't want. You subract cases you don't want. Most often, this will double subract some cases, so add them back. etc. You get a start, subractions, additions (enough for this case), subractions, additions, etc. Generally, without enormously fewer terms than your post 6 you can compute solutions to complex problems of this type.

The solution I outline invovles a total of 4 combinations you must actually compute, followed by 5 arithmetic operations.

 

[PAllen]

Think about the following:

17c6 - 10c6 - 11c6 -13c6 +1 +7

When you understand why this works, you are on your way to efficiently solving such problems.

 

[tiny-tim]

hi cloud360!

(just got up :zzz: …)

PAllen's method is definitely quicker (and therefore better) … it only involves calculating five combinations, instead of ten

 

[cloud360]

'missing all physicists' is one possibility (e.g. containing only chemists and biologists); missing all biologists means containing only chemists and physicists; I hoped that would be clear.

Note, separately, it is possible to deduce that the final answer is an even number, so 9997 cannot be correct.

how do u know the answer must be even. please can you tell me?

it is something very helpful to know

 

[cloud360]

Think about the following:

17c6 - 10c6 - 11c6 -13c6 +1 +7

When you understand why this works, you are on your way to efficiently solving such problems.

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

9988+1+7=9996

Why +1, and +7?

I know 10c6 , 11c6,13c6 are the cases where their is NOT at least 1 of each

 

[PAllen]

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

9988+1+7=9996

Why +1, and +7?

I know 10c6 , 11c6,13c6 are the cases where their is NOT at least 1 of each

Think about this statement from my first post:

But these subtractions doubly remove committees missing both physicists and chemists; and also missing both physicists and biologists. Compensate for this.

-----
Note, it is very worthwhile understanding the general method here. Instead of only adding contributions of specific non-overlapping cases, you alternate addition and subtraction of smaller and smaller corrections. Using only addition requires precise enumeration of all cases (as here, the 10 detailed cases you correctly computed and then made only a tiny addition mistake. Using alternating corrections avoids having to list all cases. In big problems of this type it can be the difference between dozens versus thousands of terms.

 

[cloud360]

Think about this statement from my first post:

But these subtractions doubly remove committees missing both physicists and chemists; and also missing both physicists and biologists. Compensate for this.

-----
Note, it is very worthwhile understanding the general method here. Instead of only adding contributions of specific non-overlapping cases, you alternate addition and subtraction of smaller and smaller corrections. Using only addition requires precise enumeration of all cases (as here, the 10 detailed cases you correctly computed and then made only a tiny addition mistake. Using alternating corrections avoids having to list all cases. In big problems of this type it can be the difference between dozens versus thousands of terms.

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 don't want) = those you do want

those you don't want are usually a smaller group and easy to work out using this method.


but i want to know how you knew you have to add 7 and 1 to "those you don't want"

i just can't see. what does 7 and 1 represent. 7 physicists?

 

[cloud360]

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)

 

[PAllen]

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 don't want) = those you do want

those you don't want are usually a smaller group and easy to work out using this method.


but i want to know how you knew you have to add 7 and 1 to "those you don't want"

i just can't 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.

 

[cloud360]

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- $$\neg$$ 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)

 

[PAllen]

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]

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.

 

[cloud360]

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 can't identify any overlaps (i can't see why, the above combinations has got something we repeated already), please help if you have time

 

[cloud360]

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)

 

[PAllen]

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).

 

[cloud360]

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 don't we subtract 7+4...why is it 7+1?

or am i thinking of it wrong

 

[PAllen]

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 don't 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.