How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3? How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3?
If I find this by method of complementation then 
Total number of arrangements without any restrictions =$\frac{7!}{4!×2!}$. 
Then I should subtract total no of arrangements where 2 is beside either three
That means arrangements with 233,323,332 should be subtracted. if we take 233,323,332 as 3 possibilities then how should I proceed further
Please correct me if I am wrong here.
 A: If $2$ is at the left end, or at the right end of the permutation, we can choose $2$ of $5$ spots where we put the the $3$'s, which can be done in ${5}\choose{2}$ ways. If $2$ is in one of the middle $5$ spots, then we can put the $3$'s in $2$ of $4$ different spots, so we can do this in ${4}\choose{2}$ ways. This should give the answer $$2 \cdot {{5}\choose{2}} + 5 \cdot {{4}\choose{2}} = 50.$$
A: As you observed, there are 
$$\binom{7}{4}\binom{3}{2}\binom{1}{1} = \frac{7!}{4!2!1!} = \frac{7!}{4!2!}$$
distinguishable arrangements of four $1$s, one $2$, and two $3$s.  From these, we must exclude those arrangements in which a $3$ is adjacent to the $2$.
A $3$ is adjacent to the $2$:  We have six objects to arrange, four $1$s, a $3$, and a block containing a $2$ and a $3$.  We choose four of the six positions for the $1$s, one of the remaining two positions for the $3$, and the final position for the block.  The $2$ and $3$ can be arranged within the block in $2!$ ways.  Hence, the number of such arrangements is 
$$\binom{6}{4}\binom{2}{1}\binom{1}{1} \cdot 2! = \frac{6!}{4!1!1!} \cdot 2! = \frac{6!2!}{4!}$$
However, in counting arrangements in which a $3$ is adjacent to the $2$, we have counted arrangements in which both $3$s are adjacent to the $2$ twice, once when we designated $32$ as our block and once when we designated $23$ as our block.  We only want to subtract such arrangements once.  Therefore, we must add them back.
Both $3$s are adjacent to the $2$:  This can only occur if the arrangement includes the block $323$.  Thus, we have five objects to arrange, the four $1$s and the block.  We choose four of those five positions for the $1$s.  The block must be placed in the remaining position.  Hence, there are 
$$\binom{5}{4} = \frac{5!}{4!}$$
such arrangements.  
By the Inclusion-Exclusion Principle, there are 
$$\binom{7}{4}\binom{3}{2}\binom{1}{1} - \binom{6}{4}\binom{2}{1}\binom{1}{1}2! + \binom{5}{4}\binom{1}{1} = \frac{7!}{4!2!} - \frac{6!2!}{4!} + \frac{5!}{4!} = 50$$
permissible arrangements.
A: Now that we have established that the three possibilities. We check how many ways we can place them into the rest of your list 
$$\_1\_1\_1\_1\_$$
Note how there are $5$ different places to place your three possibilities. That means in total you have $15$ permutations with the 2 next to a 3.
EDIT:
As mentioned in the comments, this is incorrect. We did not take into account the possibility of only $23$ occurring in a slot. There are 2 ways to arrange $23$ with 5 places to put them in. Then there are 4 free spaces. So in fact we have the $15$ permutations above with the $2(4)(5)$ permutations we found now which fives us a total of $55$ permutations.
