Choosing a Dissertation Committee: Combination Permutation Dilemma

[elove]

I am in a group we cannot agree on how to answer the following word problem...

"Let me tell you a little dirty secret about the mutual dislike between mathematicians and physicists. It can escalate into a full blown war if diplomacy is not attempted properly.

It happens that a graduate student in theoretical/mathematical physics is looking for five members of his dissertation committee. He has been working closely with three professors in the mathematics department, and 5 professors in the physics department on his dissertation research.

Now comes the monkey wrench.

The mathematics department demands that the chair of the dissertation committee must be a mathematician in order to keep the physicists in check. What? How arrogant! But, there is no other choice if a dissertation committee has to be assembled in time. As usual, physicists have to swallow their pride in order to keep peace. Yes, they are just a bunch of you know what (click on the link ) in the mathematics department!

For comic relief, can you figure out how may ways can this hapless graduate student can choose among his beloved professors if the chair of the committee must be a mathematician, and the rest of the committee can be a mix of mathematicians and physicists?"

one of the group members suggested this formula:

8C3 = 8! / (8 - 5)! x 5!

40,320 / 5! x 3!

40,320 / 120 x 6

40,320 / 720

56

but then another member said no that we need to calculate all the 4-man and 3-man combinations and add them all together

here is what i thought:

__ x__ x__ x__ x__ x each space represents a position on the committee

the chair of the committee must be a mathematician and there are only 3 options. so...

3 x __ x__ x__ x__ then the rest of the spaces represents the possible number of people to fill that position

3 x 7 x 6 x 5 x 4 i started at 7 because 1 of the mathematicians is out

this = 2,520

8! / 2,520 = 16 but now that I've typed this all out 16 sounds too low maybe?

HELP! we can't agree and are WAY off

 

[Jameson]

Hi elove,

I agree that the first slot would be 3 because there are 3 possible choices for the chair. Then the question is how many ways can you arrange the remaining members in a group of 4, where order doesn't matter?

 

[elove]

okay so I seem to be on the right track...the first ___ is 3 and since there are 4 more positions I started with 7 since there are 7 out of 8 possibilities left.

so 3 x 7 x 6 x 5 x 4 = 2,520

BUT maybe you start at 5? and not 7? maybe start with 5 since I am giving 3 possibilities out of 8.

3 x 5 x 4 x 3 x 2 = 360

that seems like too many combinations to be correct.

 

[MarkFL]

...the first ___ is 3 and since there are 4 more positions I started with 7 since there are 7 out of 8 possibilities left.

so 3 x 7 x 6 x 5 x 4 = 2,520...

You are on the right track here...and what you have can be written as:

\(\displaystyle N=3\frac{7!}{3!}=3\cdot\,_{7}{P}_{3}\)

However, you have made the order of the remaining 4 members matter (permutations), and as Jameson pointed out, order doesn't matter, so how can you modify your formula so that order doesn't matter?

 

[elove]

ahhh, I see what you are saying.

Geez, I haven't taken statistics since high school.

I don't know how to make the order not matter. My online class group is not helping.

 

[MarkFL]

ahhh, I see what you are saying.

Geez, I haven't taken statistics since high school.

I don't know how to make the order not matter. My online class group is not helping.

How many ways can we order the 4 remaining members of the committee? If we can figure out how many ways to order 4 distinct objects, and divide our result by this number, then instead of finding the permutations of 7 things taken 4 at a time, we can find the number of ways of choosing 4 things from 7...and being able to find the number of ways to choose $r$ from $n$ is very useful in statistics/probability. :)

 

[elove]

ok so

n! / r! (n-r)!

n = 7
r = 4

7! / (4! x (7 - 4)!))

7! / (4! x 3!)

5,040 / 24 x 6

5,040 / 144

= 35

but is this considering the fact that more than one mathematician can be on the committee?

to me it seems as though this formula only factors one mathematician being on the committee.

so do i 35 x 3 (to factor the other mathematicians?)

 

[MarkFL]

Yes, you have it exactly right, you want:

\(\displaystyle N=3{7 \choose 4}=3\cdot\frac{7!}{4!(7-4)!}=105\)

 

[elove]

Yes, you have it exactly right, you want:

\(\displaystyle N=3{7 \choose 3}=3\cdot\frac{7!}{3!(7-3)!}=105\)

Thank you!

 

[MarkFL]

Thank you!

I have corrected my post (post #8)...I inadvertently wrote the wrong number being chose from 7, however because of the symmetry of the binomial coefficients, the end result was still correct.

The symmetry I am referring to is:

\(\displaystyle {n \choose r}={n \choose n-r}\)

 

[elove]

i was just about to ask that... thank you for clarifying. r = 4, not 3. but the end result is still the same.

 

[elove]

but how do I know that order doesn't matter?

if order does matter than the final answer would be more than 105?

If I gave them letters to represent members...

ABCDE

is the same as

BACDE - its the same members, just a different order of the combination.

if order mattered this is 2 possible committees, if order doesn't matter this is just 1 possible committee made up of the same members.

I am confident in the formula you guided me to but I am second guessing the assumption that order doesn't matter.

 

[MarkFL]

Yes, if we let C be the committee chair, and the W, X, Y, Z be the other four committee members, then we see that:

CWXYZ

is the same as:

CZYXW

We only need be concerned with the fact that there are 5 members, one of which is the chair. The order in which they are picked doesn't matter, so long as we pick the chair first to make certain we have a mathematician to choose for that position. It is the same committee regardless of the order in which the remaining 4 members are chosen. If we were to give a title to each member, then order would matter.

 

[elove]

Yes, if we let C be the committee chair, and the W, X, Y, Z be the other four committee members, then we see that:

CWXYZ

is the same as:

CZYXW

We only need be concerned with the fact that there are 5 members, one of which is the chair. The order in which they are picked doesn't matter, so long as we pick the chair first to make certain we have a mathematician to choose for that position. It is the same committee regardless of the order in which the remaining 4 members are chosen. If we were to give a title to each member, then order would matter.

okay, I follow now. I agree and was editing my last post as you responded. I agree now that order does not matter.