Probability

[Shamoon713]

How many different linear arrangements are there of the digits 1, 2, 3, 4, 5, 6 for which:
(a) 5 and 6 are next to each other ?
(b) 5 is before 6?
(c) 3 is before 4 and 5 is before 6?
(d) 3 and 4 are next to each other and 5 and 6 are also next to each other?
(e) 6 is not last in line?

[jim360]

Combinatorics isn't my strong suit, but:

(a) can have (56) or (65) in one of five different positions, and remaining 4 numbers can be any way you like, making for a total of 2*5*4! = 240 cases.

(b) if you put 6 in position 1 then there are no cases; in position 2 then you must have 56.... (4! cases);
in position 3 then 5.6... or .56... works (2*4!); in position 4 then 5 has three slots to choose from (3*4!), etc, for a total of (1+2+3+4+5)*4! = 360. The fact that this is exactly half of the available set of permutations suggests that you could also make an argument from probability/symmetry, ie something like P(5 before 6) = P(6 before 5) = 1/2 (because there is no reason to prefer 5 or 6). But I thought of the exhaustive method first.

[jim360]

Haven't answered (c-e) yet, but hopefully the ideas above will get you started. (c,d) look trickier to resolve than (b,a) but it looks like it will end up being arguable in more or less the same way.

[Old_Geezer]

I must have it wrong. I figured there were 2^3, i.e. 8 combinations of 1 to 4; multiplied by 5 locations 5/6 could be placed, multiplied by 5 locations 6/5 might be. 8x25 is 200. Where have I lost 40 ?

[jim360]

I'm afraid that's wrong from start to finish, OG. Permutations are factorial: 4 places to place 4, 3 to place 3 and so on, for 4! (not 2^3 = 8). Then you can have (56) OR (65) in 5 locations each, so you add these.

[Old_Geezer]

Ah, ok. Thanks. I knew something was up.