First, how many positions are there total: if you have six choices for
chair, there are five remaining choices for secretary, and thus four
people to choose the treasurer from, so there are 6x5x4=120 total
possibilities.
1. how many where "both B and F are secretary and treasurer".
if B is secretary and F is treasurer, the only remaining choice is
which of the remaining four should be chair, so that's four
selections. You could also have B treasurer and F secretary, so I
think the answer you're looking for is 4x2 = 8.
2. if D is chair, then 5 possibilities for sec, 4 for treasurer = 5x4 = 20.
if "D is not Secr or Treasurer" (and not chair, since we've counted
those above) then the chair can be one of the other five, the
secretary can be one of four (not D, and not the chair), the treasurer
one of three (not D, the chair, or the secretary), so there are 5x4x3
= 60 possiblities.
so, the answer is 20+60 = 80.
3. if you have eight people, and you want to choose a team of two,
there are 8x7/2 = 28 possible teams of two. (the /2 is because unlike
the previous questions, you don't count a team of A&B separately from
a team of B&A.)
then, after choosing the first team of two, you have six people
remaining, and thus 6x5/2 = 15 possible ways to choose a second team.
so the number of ways of choosing two teams of two out of the group of
eight is 28x15/2 = 210. (again, the /2 assumes that you don't
distinguish between the two teams of two; if a solution is considered
different if you switch the two teams of two, then the answer would be
420.)
if you're looking for a general discussion of these types of problems,
I found one at http://www.themathpage.com/aPreCalc/permutations-combinations.htm,
by searching google for "permutations combinations counting".
I hope this all makes sense; I'm sorry that I can't give a definitive
answer to some questions without clarifications!
--David |
