Possible arrangements of the word ARRANGEMENT In how many the alphabets of the word ARRANGEMENT can be arranged such that :

*

*Exactly two pairs of consecutive identical alphabets appear in the arrangement?

*At least two pairs of consecutive identical alphabets appear in the arrangement?


What I've tried so far:
The alphabets that can be repeated should appear more than once in the given word, these are $$\text{AA,EE,NN,RR}$$
The answer to the first part is given by :
$$\binom{4}{2}\left[\frac{9!}{\left(2!\right)^{2}}-\binom{2}{1}\frac{8!}{\left(2!\right)^{1}}+\binom{2}{2}\frac{7!}{\left(2!\right)^{0}}\right]$$
For the second question it's needed to subtract the cases where less than two pair of consecutive alphabets appear in the arrangement from the total arrangements of the given word:
$$\frac{11!}{\left(2!\right)^{4}}$$
$$-\left[\underbrace{\frac{11!}{\left(2!\right)^{4}}-\binom{4}{1}\frac{10!}{\left(2!\right)^{3}}+\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}-\binom{4}{3}\frac{8!}{\left(2!\right)^{1}}+\binom{4}{4}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with no pair of consecutive identical alphabets}}\right]$$
$$-\binom{4}{1}\left[\underbrace{\frac{10!}{\left(2!\right)^{3}}-\binom{3}{1}\frac{9!}{\left(2!\right)^{2}}+\binom{3}{2}\frac{8!}{\left(2!\right)^{1}}-\binom{3}{3}\frac{7!}{\left(2!\right)^{0}}}_{\text{Arrangements with only one pair of consecutive identical alphabets}}\right]$$
$$=\color{green}{398160}$$
But the other way I came with was to compute the number of arrangements with two pairs of consecutive identical alphabets which is given by $$\binom{4}{2}\frac{9!}{\left(2!\right)^{2}}$$
I thought that these are the arrangements that at least two such pairs will appear but I don't understand why this way gives the wrong answer?
 A: On your second approach for at least two pairs of consecutive identical letters,
Yes $\binom{4}{2} \frac{9!}{\left(2!\right)^{2}}$ contains all arrangements of at least two doubles. The problem is that it overcounts them.
For any two given pairs (say $AA, EE$), number of arrangements is $\frac{9!}{(2!)^{2}}$ and this contains all arrangements of triples of $(AA, EE, NN)$ and $(AA, EE, RR)$ - two possible triples for the given two pairs. But when your given two pairs are (say $EE, NN$), your arrangements again include triples of $(AA, EE, RR)$.
In short there are only $\binom{4}{3}$ ways of choosing at least $3$ pairs of two identical letters but you are counting them $\binom{4}{2} \times 2 \,$ times. So you need to subtract those extra $8$ times.
Similarly for arrangements of $4$ pairs, there is only one way to choose them. But each of the $\binom{4}{2} = 6 \,$ given two pairs counted them once. Then when we subtracted $8$ times of at least three pairs, we subtracted them $8$ times and so we are net $(-2)$ on arrangements of $4$ pairs. We need to add $3$ arrangements of $4$ pairs to make sure we are counting all $4$ pairs once. Hence our answer should be -
$\displaystyle \binom{4}{2} \frac{9!}{\left(2!\right)^{2}} - 2 \times \binom{4}{3} \frac{8!}{\left(2!\right)^{1}} + 3 \times \binom {4}{4} \frac{7!}{\left(2!\right)^{0}}$
A: An easier way to get the answer to the second part, having already computed $n(exactly\; 2\;pairs) = 332,640 \;$would be to add $n(exactly\; 3\;pairs) + n(exactly\; 4\;pairs)\\
= \binom 43\left[\frac{8!}{(2!)^1}-\binom 4 4 \cdot\frac{7!}{(2!)^0}\right] + \binom 4 4\cdot \frac{7!}{(2!)^0}$
This would also have maintained the inclusion-exclusion formula used in the same pattern as in the first part.
