# Maximum number of weeks that can elapse without having the same four individuals who have served together with the same committee chair

"Each week, a subcommittee of four individuals is formed from among the members of a committee comprising seven individuals. Two subcommittee members are then assigned to lead the subcommittee, one as chair and the other as secretary.

Calculate the maximum number of consecutive weeks that can elapse without having the subcommittee contain four individuals who have previously served together with the same subcommittee chair."

According to the solutions manual, the answer should be $$140$$ weeks, however I'm getting $$836$$ weeks.

My approach was:
(1) compute binomial coefficients for both segments ($$\binom{7}{4}$$, and $$\binom{4}{2}$$)
(2) compute their product (i.e. Fundamental Principle of Counting) and I get $$210$$
(3) Therefore, unique $$4$$-person combo occurs once in $$210$$ weeks, i.e. number of elapsed weeks between re-occurrences $$= 210-1 = 209$$
(4) within that unique group, probability of a particular member becoming Chair is $$1/4$$, therefore it will take FOUR times as long to see a re-occurrence of that same person as Chair
(5) $$209$$ weeks $$\cdot 4 = 836$$ weeks

Does anybody have any ideas how to solve this problem?

• Suggestion: The secretary is irrelevant, so just ignore him. Since the chair is special, try first picking the chair and then picking the remaining members. – awkward Sep 26 '18 at 18:22
• Does it help you to notice that $7{6\choose 3} = 140?$ – BruceET Sep 26 '18 at 20:14
• Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site. – N. F. Taussig Sep 27 '18 at 11:31
• Thnx for the info and edits. – Oke Uwechue Oct 8 '18 at 0:05

We are asked to distinguish committees by which four people serve on the committee and who among those four people serves as the chairperson. The maximum number of weeks that may elapse without repetition is equal to the number of such committees, which is $$\binom{7}{4}\binom{4}{1} = 140$$ since we must choose four of the seven people to serve on the committee and select one of those four people to serve as the chairperson.

What did you do wrong?

The number of four-person committees is $$\binom{7}{4} = 35$$
Your count $$\binom{7}{4}\binom{4}{2} = 210$$ is the number of four-person committees with two officers selected from within the committee. If you were trying to count the number of four-person committees with a chairperson and a secretary, you would get $$\binom{7}{4}\binom{4}{2}2! = 420$$ since you have to assign one of those officers to be the chairperson and the other to be the secretary.

As stated above, what we are interested in is the number of four-person committees with a chairperson selected from within the committee, which is $$\binom{7}{4}\binom{4}{1} = 140$$

The number of weeks until a committee must be repeated is found by adding one to the number of ways different committees that can be formed. Thus, it would take $$141$$ weeks to ensure that some four-person committee with the same chairperson would be selected.

• Thnx. Will look through this solution – Oke Uwechue Sep 27 '18 at 16:44