Hi all
Pls check the link and figur 1.3, the right side. We're discussing with a few friends on how they calculated the variations (1,65, 7365). We are no experts at all, but we think we're missing something here, based on the reading of the explanation of this case. Any idea?
Source: Model bAsed testing essentials -anne kramer p 18
Tnx!
Ok first part I already solved.
1: if nobody is permitted to send a message, then only variation is the one where no messages are sent
65: 4 + 12 + 24 + 24 = 64 + one idle (permutation)
1chat: 4 variations
2 chats: 4!/(4-2)! = 12
3 4x3x2 =24
4 4x3x2x1 = 24
In total 64 + 1 idle: 65
Now the next number..