Let me call those names A,B,C,D... R,S,T
It's hard to go directly after the question. Why not consider finding how many teams could be built if the team contains at least 2 students whose surnames listed consecutively?
Case 1:In the team, all 3 students' names are listed consecutively.
For example: ABC, BCD, CDE
There are 20 - 2 = 18
Case 2: Only 2 students' names are listed consecutively.
For example: ABD, ABE, ABF, ... BCF, BCG, BCH...
There are 20-1 = 19 consecutive pairs starting from AB, BC, CD to ST.
If i started with CD for example, then the third student can't be B or E. I have only 20 - 2 - 2 = 16 choices for the third student.
However, this is only true for pairs from BC to RS. I will treat the pairs AB and ST separately.
Then there are 17 * 16 = 272 ways to build the team that contains a consecutive pair from BC to RS
AB can be combined with anything from D to T, 20-2-1=17 letters, to complete the team. The ST case is similar.
Then there are 2 * 17 = 34 ways to build the team that contains AB or ST.
Add up all the cases.
The number of ways that the team could be built if the team contains at least 2 students whose surnames listed consecutively is:
18 + 272 + 34 = 324
Building the team randomly: 20C3 = 20!/3!/17! = 20*19*18/6 = 1140
The final answer is 1140 - 324 = 816
Can someone confirm this?
