(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.