In how many ways In how many ways can $n$ people split in three groups and then people in each group arrange in row.
I need help to solve this. I tried to solve this in following way
$\dfrac{1}{3!}\binom{n}{n_1,n_2,n_3}f_{n_1}f_{n_2}f_{n_3}$, where  $f_{n_i}=n_i! $ is number ways that set of $n$ people arrange in row.
 A: My initial guess is that it would be the number of ways you could arrange $n$ people in a row multiplied by the number ways you could then drop two dividers into the spaces between them ${n-1}\choose2$.  So $$n!{{n-1}\choose2}$$
That solution assumes each group is different (there is an A group, a B group and a C group).  Like @JMoravitz asked in a comment. 
If the groups are all equivalent it would be $$\frac{n!{{n-1}\choose2}}{3!}$$since 3 groups can be arranged $3!$ ways, all of which would really be the same result.
Editing to add:If empty set groups are allowed, each divider has $n+1$ places it can be dropped.  Repeats allowed.
A: This doesn't matter if you divide them into groups and then permute, the result is same as long as final sitting pattern is concerned, you cannot distinguish them from your method hence total ways would be:
$$\newcommand{\c}[2]{{}^{#1}{\mathbb C}_{#2}}
\sum_{k+l+m=n} \frac{\c {k+l+m}k\c{l+m}l\c{m}{m}}{3!}3!k!l!m!=n!$$
Now what you are seeking is:
$$\sum_{k+l+m=n} \frac{\c {k+l+m}k\c{l+m}l\c{m}{m}}{3!}k!l!m!=\frac{n!}{3!}$$
