Answer
(a) $48$
(b) $60$
(c) $72$
Work Step by Step
(a)A and B must sit together;
We take A and B as one unit F, for C D E F there are $4!=24 ways$
As A and B can exchanges seats, the total would be $24\times2=48$
(b) C must sit to the right of, but not necessarily next to, B;
We let B sit first, B can sit on the left 4 seats:
B1: total $4\times3\times2=24$ ways for the rest to seat
B2: first seat has 3 choices (A D E), total $3\times3\times2=18$ ways
B3: total $3\times2\times2=12$ ways
B4: total $3\times2=6$ ways
In total, there are $24+18+12+6=60$ ways.
(c) D and E will not sit next to each other?
We calculated in (a) for A and B together, it is the same for D and E together (48)
we can subtract this number from the full permutation, so
$5!-48=72$ is the answer.
