What is the number of arrangements of 5 different things taken 3 at a time in which 2 particular thing always occur? I know , there is formula to apply, s!(r-s+1)( n-s P k-s).

I tried with example, 
n=5 ; r=3 and s=2. 
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.

But, if I solve manually with example, 
consider n = {1,2,3,4,5}
then possible arrangements are as follows:

145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18. 
Where am I going wrong
The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.

The formula that deals with what you want is $r!{n-s \choose r-s}$. In this case, the number would be $3!{3 \choose 1}=18$, exactly what you got.
2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18