Problem : Permutation and Combination : In how many ways can we divide 12 students in groups of fours. Selecting groups of r items from n items
Example : In how many ways we can select groups of two from 6
students. 
Concept : 
Let the students be A, B, C, D, E, F . A can group with
each of the remaining five as AB, AC, AD, AE, AF now
we have B, C, D, E, F left, B can group with each of
the remaining four as BC, BD, BE, BF, now we have C,
D, E, F left, C can group with each of the remaining
three as CD, CE, CF, now we have D, E, F left, D can
group with each of the remaining two as DE and DF, E
can group with the last one F , EF. Therefore we have
AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE,
DF and EF, fifteen groups of twos.
Alternatively
One student A can select the other one of his group in
$^5C_1$ ways . If B is another student who does not belong
in the group A, we can select the other one of his group
$^3C_1$ ways the remaining two consists of the third group.
= 5 $\times 3 \times 1$
But as per my understanding : 
I used : $^6C_2$ which is equal to 15. 
However if I am using this concept in the following question  : 
Problem : 
In how many ways can we divide 12 students in groups
of fours.
A group of 4 can be chosen from 12 students is : 
$^4C_{12} = 495$ which is wrong answer and answer is $5775$ 
Please help on this.. Thanks..

Perhaps it is easier to see the general case, having to divide $nk$ students in groups of $k$. One way to do it is the following, although the approach of Benjamin Dickman (which lends itself to a straightforward generalization) is probably cleaner.
Pick a student at random, and "pair" her with the
$$
\binom{nk-1}{k-1}
$$
groups of $k-1$ among the other $nk-1$.
Now you are left with $(n-1)k$ students. Pick one at random, and pair him with the 
$$
\binom{(n-1)k-1}{k-1}
$$
groups of $k-1$ among the other $(n-1)k-1$.
At this point you will have guessed that the formula is
$$
\prod_{i=0}^{n-1} \binom{(n-i)k-1}{k-1}.
$$
(The last factor is $1$, but I have included it for completeness.)
Benjamin Dickman's formula would be the equivalent
$$
\frac{(nk)!}{n! (k!)^{n}}.
$$
In your example of the $12$ students to be divided in groups of $4$, you have $n = 3$ and $k = 4$, and this reads
$$
\binom{11}{3} \cdot \binom{7}{3} \cdot \binom{3}{3}
=
165 \cdot 35 \cdot 1
=
5775.
$$
