Circular P&C

may be very bad a method but i like this...donno if it fits in here but have a look...

first of all i ll solve the problem in which it is not circular permutation but linear one...and then try to make it circular...[1][1][1]

let the train stop at 3 stations such that s₁<s₂<s₃

now given condition, the trains cannot stop in adjacent station...

so, from the problem we have..
1≤s₁
s₁+2≤s₂
s₂+2≤s₃
s₃≤10

but its like hell if we try to solve these inequations and try to find the no of solutions (s₁,s₂,s₃)
so instead we ll convert these to an equation...

so they yeild to,

1+a=s₁
s₁+2+b=s₂
s₂+2+c=s₃
s₃+d=10
where a,b,c,dε[0,∞)

adding the above four equations we get,

1+a+2+b+2+c+d=10
or, a+b+c+d=5

now trying to find no of solutions of s₁,s₂,s₃ from the inequations is equivalent to finding possible set of value of a,b,c and d...

we have to partition the no 6 to give 4 nos using 3 partitions
so no of ways is ⁸C₃

so the train can stop in ⁸C₃ ways...[1][1][1]

if i did not go wrong anywhere..

neways this was the soln if the station were nt in circular arrangement...so now i ll try to delve into the actual problem!!

two cases...frst wen train stops at 1 and second the train doesnt

wen train stops at 1

let s₁=1
so next it has to stop newhere from station 3 to station 9
in two positions,

so
3≤s₂
s₂+2≤s₃
s₃≤9

whch boils down to
a+b+c=4.........[jumping many steps[3][3]]

no of ways is ⁶C₂..

wen train doesnt stop in 1

its a linear permutation in which all the three stations will lie within 2 to 10

so no of ways after solving this will be

⁷C₃ most probably...

so total no of ways is ⁶C₂+⁷C₃=15+35=50

and i am completely unsure wether or not answer is correct!!
some expert please help!

does someone have an easier method???

Good work all of you :)