Question

There are six students A,B,C,D,E,F.
In how many ways can they be seated -in a line so that C and D do not sit together ?

Answer 1

Total-no, of ways = 6! = 720. Treat C and D as one and we have 6 2 + 1 = 5 units which can be arranged in 5! = 120 ways. But C and D can be arranged in 2! = 2 ways. Hence the no. of ways when they are together is 120 $\times$ 2 = 240.
Hence the arrangements when C and D will not be together is 720 - 240 = 480.
Alternative Method:
Ignore C and D. The rest four can be arrranged in 4! = 24ways. There will be 5 gaps iii between each arrangement of these 4 students. In these 5 gaps two C and P can be arranged in
${}^6{P}_2$ = 5 $\times$ 4 = 20 ways. Hence the total number of ways when C and D are together is 24 $\times$ 20 = 480.