Correction factor In permutation and combination So It's always confusing for me where to apply the correction factor in permutation and combination. For example there is a question like find the number of ways in which 4 distinct object can be divided into 2 groups such that one group has 2 objects and other group has 2 objects. Can someone explain me how to apply correction factor to this question
Thanks.
 A: It depends on whether the groups are labelled or unlabeled
Suppose you have $4$ objects, $A,B,C,D$ to be divided into two equal groups.
The first group can be formed in $\binom42$ = $6$ ways
Now if the groups are labelled, say East and West, there will actually be $6$ possible groups of two, but see what happens when they are unlabeled !
$AB-CD$ becomes the same as $CD-AB$, so you need to divide the result by $2$, or more correctly, by $2!$
If you wanted to form three groups of $2$ from $6$ objects, the number of unlabeled groups would be $[\binom62\binom42\binom22] \div 3!$
Note that groups can get automatically labelled by differing size, or differing type (e.g. gender)

Response to OP's comment
Labelled = named, or distinguishable
Unlabeled = unnamed, or indistinguishable
So if from $4$ objects you make two groups of two, if the groups are labeled Eat and West, groups are
EAST WEST
$AB\quad CD$
$AC\quad BD$
$AD\quad BC$
$BC\quad AD$
$BD\quad AD$
$CD\quad AB$ .... $6$ labelled groups
But if they are unlabeled, how can we distinguish between $AB\;CD$ and $CD\;AB$ and so on, so now only three unlabeled groups can be formed,
$AB\quad CD$
$AC\quad BD\;$
$AD\quad BC$
A: Assuming that the two groups are to be regarded as indistinguishable, the enumeration of $~\displaystyle \binom{4}{2} = 6~$ is wrong.
Consider the grouping:

*

*Group-1 : Person-1, Person-2

*Group-2 : Person-3, Person-4

Now, consider the associated grouping:

*

*Group-1 : Person-3, Person-4

*Group-2 : Person-1, Person-2

If the two groupings are to be considered the same, then you have a problem, because in the $~\displaystyle \binom{4}{2}~$ enumeration, they are counted twice.  You will have the same over-counting for any other grouping.
So, the correct enumeration, when Group-1 and Group-2 are to be regarded as indistinguishable from each other is
$$\binom{4}{2} \times \frac{1}{2!}.$$

An alternative viewpoint is to ask, who will Person-1 be grouped with.  There are only $3$ choices, and once Person-1's partner has been chosen, the groupings are set.
This assumes that it is irrelevant whether Person-1 and their partner are chosen in Group-1 or Group-2.
A: The multiplication principle applies when counting lists, and lists have a distinct order to them. But sometimes we are counting objects for which certain orderings/rearrangements aren't important. In these situations, we divide the total number of lists by the number of rearrangements which are to be considered “equivalent."
In your example, let the four objects be numbered as 1, 2, 3, 4. To partition them into two sets of two elements, we can:

*

*start with a list of the four objects

*take the first pair and assign them to the first set

*take the second pair and assign them to the second set.

So if the list is $(1,4,3,2)$, the associated partition is $\{\{1,4\},\{3,2\}\}$.
There are $4! = 24$ lists of the four objects, so you have at most $24$ partitions. However, this process will result in many lists generating the same partition! For instance, $(4,1,3,2)$, $(1,4,2,3)$, and $(2,3,4,1)$ will all generate the same partition as $(1,4,3,2)$. This is because lists are ordered and sets are not.
So the next thing to count is how many lists generate the same partition. Given any list $(a,b,c,d)$, these are the lists which generate the same set $\{\{a,b\},\{c,d\}\}$:
$$\begin{array}{cccc}
    (a,b,c,d) & (b,a,c,d) & (a,b,d,c) & (b,a,d,c) \\
    (c,d,a,b) & (c,d,b,a) & (d,c,a,b) & (d,c,b,a) 
\end{array}$$
Another solution to this question: notice that there are three permutations of the list elements that don't affect the associated partition:

*

*swapping the first pair: $(a,b,c,d) \mapsto (b,a,c,d)$ [because $\{b,a\} = \{a,b\}$, so $\{\{b,a\},\{c,d\}\} = \{\{a,b\},\{c,d\}\}$]

*swapping the second pair: $(a,b,c,d) \mapsto (a,b,d,c)$

*swapping the two pairs: $(a,b,c,d) \mapsto (c,d,a,b)$
We can do any combination of these three permutations too. Swap first and second, swap first then the two pairs, all three swaps, etc.
So the number of lists equivalent to any one list (including the one we started with) is $2 \times 2 \times 2 = 8$.
Putting this together: the $24$ lists are divided into groups of $8$. Each group corresponds to one and only one partition. So the number of partitions equals the number of groups. That is, $24 \div 8 = 3$.
If you let go of “permutations and combinations” and think instead about equivalences on sets of lists, you can solve more problems.
