Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? OA 40

Let, Michael and Anthony be considered as a single unit. Then we have to select 1 person from remaining 4 to form a 3 member subcommitee. 1 person can be selected in 4C1 =4 ways. Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8 Total no of ways of selecting 3 people from 6 = 6C3=20 Percentage = 8/20 = 40%

jimmiejaz wrote:Let, Michael and Anthony be considered as a single unit. Then we have to select 1 person from remaining 4 to form a 3 member subcommitee. 1 person can be selected in 4C1 =4 ways. Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8 Total no of ways of selecting 3 people from 6 = 6C3=20 Percentage = 8/20 = 40% Beautiful.

what percent of all the possible subcommittees that include Michael also include Anthony? Total number of 3 member committees that include Michael: 5C2 = 10 (with Michael as the third member) Total number of 3 member committees that include Michael and Anthony: 4C1 = 4 (with Michael and Anthony as the two members) now 4 is what percent of 10? = 40%

jimmiejaz, logitech Could you please explain this step:- Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8 Question - Why did you do this ? Once you have Michael and Anthony together, there are just 4 different ways in which 2 subcommitees could be formed. Why multiply by 2 ? The second subcommittee is formed automatically once you have chose the third person who would be with Anthony and Michael.