I am not sure if I am doing the following problem correctly.
A six person person committee composed of A, B, C, D, E, F, is to select a chairperson, a secretary, and a treasurer. How many selections include A or B or both?
Ways for A and not B: 1 (position 1 i.e, A) * 4 (position 2) * 3 (position 3) * 3! (ways to arrange the selections in the 3 positions) = 72
Ways for B and not A: 1*4*3*3! = 72
Ways for B and A: 1*1*4*3! = 24
72 + 72 + 24 = 168
I had another thought on how to solve this problem.
nPr(6,3) (the total possible ways to select the positions) - nPr(4,3) (the ways to select the positions without a or b) = 96