Arranging “success” such that 2 “s” come together when (i) all “s” are identical (ii) all “s” are distinct 
How many ways can the letters of the word "success" be arranged such that two "s" come next together ($3$ ”s” formation is allowed) if (i) all s -es are identical (ii) if all s -es are distinct (for eg, s1,s2,s3)

My solutions :
For (i) As all s -es are identical, we don’t need to worry about the arrangement, so the answer would be $6!/2! - 5!/2!$, which is equal to $300$. The according equation is formed because there will be repetitions in $3$ “s” formations, for example (ss)s = s(ss).
For (ii), when they are distinct, they will permute themselves inside the groups, so my answer would be $(6!/2! \times 2! \times 3) - (5!/2! \times 3!)$, which is equal to $360$. Edit- I meant by [(6!/2!) x 2! x 3] - [(5!/2!) x 3!] (=1800). My bad.
I still did the minus because (s1s2)s3 would equal to s1(s2s3).
However, I am not confident about my answers, so I would be so grateful if I can get help.
 A: Your first answer is correct, but the second is too small.
(i) Let's verify your answer with a different method.
We will subtract the number of arrangements of the letters of the word $SUCCESS$ with no adjacent $S$s from the total number of arrangements.
The word $SUCCESS$ contains seven letters, so we have seven positions to fill.  We can fill three of them with $S$s in $\binom{7}{3}$ ways, two of the remaining four positions with $C$s in $\binom{4}{2}$ ways, then arrange the distinct letters $U$ and $E$ in the remaining two positions in $2!$ ways.  Hence, there are
$$\binom{7}{3}\binom{4}{2}2!$$
arrangements of the letters of the word $SUCCESS$.
To count the number of arrangements with no consecutive $S$s, we first arrange the four letters $C, C, E, U$.  There are four positions to fill.  We can fill two of them with $C$s in $\binom{4}{2}$ ways, then arrange the distinct letters $U$ and $E$ in the remaining two positions in $2!$ ways.  This creates five spaces in which to place the $S$s, three between successive letters in the arrangement of $C, C, E, U$ and two at the ends of the row.  $$\square L_1 \square L_2 \square L_3 \square L_4 \square$$
where the $L$s represent the letters $U, E, C, C$ and the squares represent the spaces where we can place an $S$.  To ensure that no two $S$s are consecutive, we choose three of these five spaces in which to place an $S$, which can be done in $\binom{5}{3}$ ways.  Thus, there are
$$\binom{4}{2}2!\binom{5}{3}$$
arrangements of the letters of the word $SUCCESS$ in which no two of the $S$s are adjacent.
Thus, the number of arrangements of the letters of the word $SUCCESS$ in which at least two $S$s are adjacent is
$$\binom{7}{3}\binom{4}{2}2! - \binom{4}{2}2!\binom{5}{3} = 300$$
as you found.
(ii) If the $S$s are distinct, then we must arrange the letters
$$S_1, S_2, S_3, C, C, E, U$$
so that no two of the $S$s are adjacent.
If there were no restrictions, we could choose two of the seven positions for the $C$s, then arrange the five distinct letters $U, E, S_1, S_2, S_3$ in $5!$ ways, giving
$$\binom{7}{2}5!$$
possible arrangements.
From these, we subtract those with at least two adjacent $S$s.  We can first arrange the four letters $C, C, E, U$, which can be done in $\binom{4}{2}2!$ ways since we must choose two of the four positions for the $C$s, then arrange the distinct letters $U$ and $E$ in the remaining two positions.  This creates five spaces in which to place the $S$s.
$$\square L_1 \square L_2 \square L_3 \square L_4 \square$$
where the $L$s represent the letters $C, C, E, U$ and the squares represent the spaces where we can place an $S$.  To ensure that no two $S$s are consecutive, we choose three of these five spaces in which to place an $S$, which can be done in $\binom{5}{3}$ ways.  The three distinct $S$s can be arranged in the selected spaces in $3!$ ways.  Hence, there are
$$\binom{4}{2}2!\binom{5}{3}3!$$
arrangements in which no two Ss are adjacent.
Therefore, there are
$$\binom{7}{2}5! - \binom{4}{2}2!\binom{5}{3}3! = 1800$$
arrangements in which at least two $S$ are adjacent.
