Difference between two counting arguments to find partitions of a set

Question

I am trying to understand the difference between these two questions.

Question 1: We are given an $n$-element set and non-negative integers $n_1,n_2,\ldots,n_r,$ whose sum is equal to $n$. The number of partitions of the set into r disjoint subsets, with the $i$th subset containing exactly $n_i$ elements is given by the multinomial coefficient $n \choose {n_1,n_2,\ldots,n_r}$

Question 2.

If we choose each of the $n_i = 2 $ in question 1. How is it different from Question 2? I am unable to see the difference between the two arguments. If possible can explain for which scenarios can i apply the answer from question 1 and for which scenarios i should not overcount. Thanks in advance

Answer 1

In question 1 we assume either that the subsets are different sizes or we care about the order they are chosen. In question 2 we do not care about what order the pairs are chosen in. If you start with $\{1,2,3,4\}$ do you consider the partition into $\{1,2\},\{3,4\}$ different from $\{3,4\},\{1,2\}?$ Question 1 does, question 2 does not. That makes question 2 divide by $2!$ because each partition is counted twice

Answer 2

$$ \eqalign{ & \left( \matrix{ 2n \cr \underbrace {2,2, \cdots ,2}_{n\,terms} \cr} \right) = {{\left( {2n} \right)!} \over {\left( {2!} \right)^{\,n} }} = \cr & = {{\left( {2n} \right)!} \over {2!\left( {2n - 2} \right)!}}{{\left( {2n - 2} \right)!} \over {2!\left( {2n - 4} \right)!}} \cdots {{\left( {2n - 2\left( {n - 1} \right)} \right)!} \over {2!\left( {2n - 2n} \right)!}} = \cr & = \left( \matrix{ 2n \cr 2 \cr} \right)\left( \matrix{ 2n - 2 \cr 2 \cr} \right) \cdots \left( \matrix{ 2 \cr 2 \cr} \right) \cr} $$