A)3 possiblities for character b/w A and B={C,D,E}----->3
2 permutations of A and B= {AB,BA}-----2
after placing one character b/w A and B ,and permuting them we have 3 elements left----->3!=6
i)block formed by A,B and one character.
ii) and iii) 2 characters left
Total=6*2*3=36
probablity=36/120=0.3
B)3 possiblities for character b/w A and B={CD,DE,CE}----->3
2 permutations of A and B= {AB,BA}-----2
after placing two character b/w A and B ,and permuting them we have 2 elements left----->2!=2
i)block formed by A,B and two characters.
ii) 1 character left
Total=2*2*3=12
probablity=12/120=0.1