<s>[INST] Overcounting when doing Counting Problems Problem:

How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?

Attempted Solution:
$$N_\text{Teams} = {12 \choose 2}{10 \choose 5}{ 5 \choose 5} = {12 \choose 2}{10 \choose 5}$$
Arriving at this solution by first picking the team of 2 and then the two teams of 5, however the provided solution requires that my answer be divided by 2 due to "Overcounting" which is briefly mentioned but not explained. How does one gain an intuition for overcounting in this problem and also for counting problems in general?
 [/INST] There's one more (cool, in my opinion) way to look at such problems. 
For each possible allocation of people into groups, let's take the same example as in the OP, there are a total of 
$$
\binom{12}{2}\binom{10}{5}\binom{5}{5} =16632
$$
allocations, such as (letters are people, numbers are groups):
$$
(1)AB|(2)CDEFG|(3)HIJKL
$$
Now, $\mathbf{each}$ such allocation $\textbf{enters the list exactly twice}$: 
$$
(1)AB|(2)CDEFG|(3)HIJKL\\
(1)AB|(2)HIJKL|(3)CDEFG
$$
There are no other $\mathbf{identical}$ allocation, i,e. no other way to produce an identical allocation. Hence, the number must be halved, or remove ecery second entry from the list: 
$$
\binom{12}{2}\binom{10}{5}\binom{5}{5} - \frac{\binom{12}{2}\binom{10}{5}\binom{5}{5}}{2} = 8316
$$
Here's another example: also 12 people, but 2 groups of 4 and 2 groups of 2. What do we have? The list size is 
$$
\binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2}
$$
Base allocation is 
$$
(1)ABCD|(2)EFGH|(3)IJ|(4)KL
$$
how many of such identical allocations? 
$$
(1)ABCD|(2)EFGH|(3)IJ|(4)KL\\
(1)ABCD|(2)EFGH|(3)KL|(4)IJ\\
(1)EFGH|(2)ABCD|(3)IJ|(4)KL\\
(1)ABCD|(2)EFGH|(3)KL|(4)IJ
$$
Hence $\frac{3}{4}$ of all allocations are repeated, so you need to multiply 
$$
\binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2} - \frac{3}{4} \times \binom{12}{4}\binom{8}{4}\binom{4}{2}\binom{2}{2} = 51975
$$
</s>