Question

Suppose a mathematics department has 4 full professors, 8 associate professors, and 3 assis- tant professors. How many ways are there to select a 4-person curriculum committee if (a) at least one full professor is chosen? (b) the committee contains at least one professor of each rank? Suppose, as well, that a 3-person recruitment committee must also be selected. How many ways are there to select both committees if (c) there are no restrictions? (d) the committees must be disjoint? (d) the committees can have at most one person in common?

Answer 1

$(a)Total~no.~of~ways~to~select~4~persons~curriculum~committee~where\\ at~least~one~full~professor=\binom{4}{1}\binom{11}{3}+\binom{4}{2}\binom{11}{2}+\binom{4}{3}\binom{11}{1}+\binom{4}{4}\\ =1035\\ (b)~Total~no.~of~ways~to~select~4~persons~curriculum~committee~where\\ at~least~one~professor~of~each~rank=\binom{4}{1}\binom{8}{1}\binom{3}{2}+\binom{4}{1}\binom{8}{2}\binom{3}{1}+\binom{4}{2}\binom{8}{1}\binom{3}{1}\\ =576\\ (c)~If~there~is~no~restriction~then~total~no.~of~ways~to~form~two~committees\\ =\binom{15}{4}\times \binom{15}{3}=621075\\ (d)~If~the~committees~are~disjoint~then~total~no.~of~ways~to~form~two~committees\\ =\binom{15}{4}\times \binom{11}{3}=225225\\ (e)~Total~no.~ways=\binom{15}{4}\times \binom{11}{3}+\binom{15}{1}\binom{14}{3}\times \binom{11}{2}=525525$