Find the number of ways to split a group of 12 into groups of three. Also, Sam and Tom won't sit together. Any strategy would help. 
Answer is $352,800$. 
My answer is correct, but it seems illogical and I don't understand it. 
Won't sit together= find all ways - Sam/Tom do sit together. 
My Answer: Won't sit together = find all ways - Sam/Tom do sit together. 
Won't sit together $= 12C3 \times 9C3 \times 6C3 \times 3C3 - 10C3 \times 7C3 \times 4C3 =352,800$
It's correct, but it's definitely not the right way to answer this. 
How can I calculate Sam/Tom sit together? 
Clarification: 12 into 4 distinct groups.
Each team A,B,C D has 3 players. 
Sam/Tom refuse to sit next to each other on any team. 
So I did find the answer, but it doesn't look right to me. 
 A: Your strategy of subtracting the number of distributions in which Sam and Tom sit together from the total number of distributions is correct.  However, the stated answer is incorrect.
As you found, the number of ways of splitting twelve people into four labeled groups of three is
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{6}{3}$$
Suppose Sam and Tom sit together.  There are four ways to choose which group they are in.  There are ten ways to choose one of the remaining people to be in the same group.  The remaining nine people can be split into three groups of three in $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ ways.  Hence, there are
$$\binom{4}{1}\binom{10}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
distributions in which Sam and Tom sit together.
Consequently, the number of admissible ways of splitting the twelve people into four labeled groups of three is
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} - \binom{4}{1}\binom{10}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3} = 302,400$$
A: My understanding to your question: total $12$ people divided into $4$ distinct group each with $3$-people. If this is true, then:
$$\textrm{Total} - \textrm{They in the same group},$$
so total:
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3},$$
and consider 'Sam-Tom-together':
$$(1\cdot 1\cdot\binom{10}{1})\binom{9}{3}\binom{6}{3}\binom{3}{3}\cdot\frac{1}{3!}\cdot4!.$$
With the power of wolfram alpha the answer is, probably, $302,400$. By the way I hope they will get together again.
