MHB Calculating Executive Committee Possibilities for a Board of 12 Directors

[alexmahone]

A company’s board of directors has 12 members. The board must select an executive committee of
any nonzero size, consisting of a president and treasurer (these titles may or may not be conferred to
the same person). How many possible ways are there to do this?

My attempt:

The answer is the number of non-zero subsets of the board of directors, which is $2^{12}-1$. Is this correct?

 

[alexmahone]

I just realized that the committee {1, 2, 3, 4, 5, 6} where 1 is the president and treasurer is probably different from the committee {1, 2, 3, 4, 5, 6} where 2 is the president and treasurer. This complicates the problem.

 

[MarkFL]

A company’s board of directors has 12 members. The board must select an executive committee of
any nonzero size, consisting of a president and treasurer (these titles may or may not be conferred to
the same person). How many possible ways are there to do this?

My attempt:

The answer is the number of non-zero subsets of the board of directors, which is $2^{12}-1$. Is this correct?

Yes, I agree if we do not consider the titles to be conferred:

$$N=\sum_{k=1}^{12}\left({12 \choose k}\right)=2^{12}-1$$

However, if we also add in the titles, I think we need:

$$N=\sum_{k=1}^{12}\left({k \choose 2}\cdot{12 \choose k}\right)=67584$$

edit: I just realized I did not account for the same person holding both titles...so perhaps it should be:

$$N=12+\sum_{k=2}^{12}\left({k \choose 2}\cdot{12 \choose k}+k\right)=67673$$

 

[alexmahone]

However, if we also add in the titles, I think we need:

$$N=\sum_{k=1}^{12}\left({k \choose 2}\cdot{12 \choose k}\right)=67584$$

Could you please explain this?

 

[Opalg]

A company’s board of directors has 12 members. The board must select an executive committee of
any nonzero size, consisting of a president and treasurer (these titles may or may not be conferred to
the same person). How many possible ways are there to do this?

My attempt:

The answer is the number of non-zero subsets of the board of directors, which is $2^{12}-1$. Is this correct?

The answer to problems of this sort often depend on how you interpret the question. In this case, are you just concerned with the choice of individuals in the committee, or does the description of the committee include specifying which individual/s have the positions of president and treasurer? If for example the committee consists of three directors, call them A, B and C, then anyone of those three could be the president, and anyone of them could be the treasurer. I would interpret that as giving nine different committees, although they all have the same set of members. But your answer counts that as just one committee.

If your interpretation of the question is correct then so is your answer. But my answer would be very different, as follows. If a single member acts as both president and treasurer then there are $$12$$ ways to choose that person, and $$2^{11}$$ possible ways to complete the committee from a subset of the remaining directors. If the president and treasurer are different, then there are $12$ ways to choose the president, $11$ ways to choose the treasurer, and $$2^{10}$$ possible ways to complete the committee. That gives a total of $$12\cdot2^{11} + 132\cdot2^{10} = 39\cdot2^{12}$$ for my answer to the question.

Edit. I somehow overlooked previous comments when posting that. I now see that the problem of how to interpret the question has already been raised.

 

[MarkFL]

Could you please explain this?

I will explain my reasoning for my latest version, added after you quoted my post:

$$N=12+\sum_{k=2}^{12}\left({k \choose 2}\cdot{12 \choose k}+k\right)=67673$$

The first term ($12$) is the number of 1 person commitees, where naturally that single person (for which we have 12 choices) holds both titles.

Now the sum begins with all the two person committees...where we first consider the number of ways to choose 2 from 12, AND the number of ways to choose 2 from 2 and then we add 2 because there are two ways for each of those 2 people to hold both titles. And the sum continues in like manner for the committees making up 3-12 people.

Does this make sense?