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Consider the problem of combining N people into 3 seats; denoted by NC3:

In tabular, the total permutation can be represented as: 

ABC  BAC  CAB  DAB  EAB  FAB  GAB  HAB  IAB  JAB.........  
ACB  BCA  CBA  DBA  EBA  FBA  GBA  HBA  TBA  JBA.........  
...  ...  ...  ...  ...  ...  ...  ...  ...  ... ........  
...  ...  ...  ...  ...  ...  ...  ...  ...  ...  .......

Rather than using the formula, taking this generally, There are N seats for the first position, N-1 for the second and N-2 for the third position. So, there are N(N-1)(N-2) possible permutations, but to make the order irrelevant(i.e. combination), we must divide it by 6?(because of the possible permutation of the three letters itself) . And this gives the correct answer, However:

For instance,
In taking ABC from the N letters(considering there are more than just 26 letters, ex: A1, there would be: ABC, ACB, BAC, BCA, CAB, CBA, permutations of it. So dividing the total possibilities would make sense. But, there are N letters, so there would be say group of DEF such 6 permutations. Shouldn't we divide the product of N(N-1)(N-2) by (3!3!)? Where does my logic fail?
EDIT: to reformulate, the 3! is for one group of objects, right? to cancel the repititions of one group,e.g. ABC, we divide it by 6, so to cancel the repetitions of 2 such groups, e.g. ABC, DEF(these have a sum total of 12 permuted outputs), shouldn't we divide by (3!3!)?

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    You just have to divide by 6 (=3*2*1) – 2017-01-06
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    To be honest I don't understand the end of your question. Can you reformulate? – 2017-01-06
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    @md5 please see my edit. – 2017-01-06

2 Answers 2

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For each set of three letters there are $6$ ways to permute them. But you only need to divide for the three letters you chose. For example in the following set:

ABC, ACB, BAC, BCA, CAB, CBA
DEF, DFE, EDF, EFD, FED, FDE

There are $12$ possibilities, $6$ with ABC and $6$ with DEF. The expected result is $2=\frac{12}{6}$, because you have to first choose the set of three letters, and afterwards cancel the permutations.

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to reformulate, the 3! is for one group of objects, right?

No, it's not. You can't attribute $3!$ to some specific group, and then say since we have $2$ groups (i.e., ABC & DEF) we need to divide by $3!3!$. I think explaining to you "why dividing by one single $3!$ is the correct approach" doesn't really solve your confusion. You need to find the flaw in your argument. As a hint, ask yourself "What happens if we divide by $3!3!$? What would be the answer if we've done so? Is it an integer? What if we had $3$ sits and $5$ people? How can you form the groups? ABC & what? DE?

Asking yourself such questions will soon reveal the problem to you.