Question

Six people (A, B, C, D, E, F) are lining up to enter a museum.

(1) How many ways are there to line up the six people if A is somewhere before B, although not necessarily immediately before?

(2) How many ways are there to line up the six people if A is next to C?

(3) How many ways are there to line up the six people if A must be next to B but C and D can not be next to each other?

Answer 1

1) Total no of ways of arrangements of 6 people = 6! = 720

Out if these ways, A and B can be arranged in 2 ways, either first A then B or either first B then A. Both the ways are equally likely. But we have to choose only first case.

Means, in every 2 ways, our favourable is only 1 case.

So no of ways = 720/2 = 360

2) If A is next to C then A and C can take place either of (1,2) or (2,3) or (3,4) or (4,5) or (5,6) in 5 ways. The other 4 can be arranged in remaining 4 places in 4! ways.

So required no of ways = 5•4! = 120

3) Case I: Let A and B are on position (1,2), then for C & D, favourable positions are (3,5) or (5,3) or (4,6) or (6,4) or (3,6) or (6,3) means 6 ways and remaining 2 can be seated in 2! = 2 ways. So no of ways = 6•2 = 12

Case II: Let A and B are on position (2,3) then for C and D, favourable positions are: (1,4),(1,5),(1,6),(4,1),(5,1),(5,6),(4,6) or (6,4) total 8 ways and remaining two can be arranged in 2 ways. No of ways = 8•2 = 16

Case III: Let A and B are on position (3,4) then for C and D, favourable positions are: (1,5),(1,6),(2,5),(2,6),(5,1),(5,2),(6,1), or (6,2) total 8 ways and remaining in 2! Ways So no of ways = 8•2! = 16

Case IV: Let A and B are on position (4,5). It's same as case II, so no of ways = 16

Case V: Let A and B are on position (5,6). It is same as case I, so no of ways = 12

Hence, total no of ways = 12+16+16+16+12 = 72