Arrangements of BANANAS where the A's are separated 
How many arrangements of the word BANANAS are there where the $3$ A's are separated?

I know that once chosen the places for the three A's, there are $\dfrac{4!}{2!}=12$ possible arrangements for the rest of the letters (we divide by $2!$ because there are $2$ N's). But I am having trouble with choosing the places for the A's.
If I do this manually, I count $10$ different arrangements for the $3$ A's, and that would mean that there is a total of $12\cdot 10$ possible arrangements that fit the initial condition. However, I would like to learn to calculate the $10$ cases with a combinatorics argument, instead of just counting. Could someone help me?
 A: In case, you want that no two $A$'s are together, then you may refer to answer by @user525966 
If you want that $3A$'s are never together (for example, BAANANS is acceptable but BAAANNS is not acceptable), then 
Imagine tying up all $3$ $A'$s by a string and consider them as one element. So effectively you have now $5$ elements ($1B,3A,N,N,1S$), which can be arranged in $5!/2!=60$ ways. And $3 A'$s amongst themselves can be arranged in $3!/3!=1$ way.
Hence, total no. of ways in which all $A'$s are together is $60$. 
Total no. of ways in which letters of BANANAS can be arranged =$\frac{7!}{2!3!}=420$ 
Total no. of ways in which all $A'$s are never together =Total$-$always together=$420-60=360$
A: Total number of linear arrangements formed from the letters $BANANAS$ (without restrictions)
$$=\frac{7!}{3!2!}$$
Let's consider two cases where $A's$ come together as follows
Case-1: Considering $\boxed{AA}$ as a single unit, the total number of linear arrangements from $A, \boxed{AA},  B, N, N, S$ (this case includes all strings with both $2A's$ and $3A's$ coming together such that strings with $\boxed{AA}A\equiv AAA$ & strings with $A\boxed{AA}\equiv AAA$ are considered as distinct but actually they aren't hence we need to remove such redundant strings)  $$=\frac{6!}{2!}$$
Case-2: Considering $\boxed{AAA}$ as a single unit, the total number of linear arrangements from $ \boxed{AAA},  B, N, N, S$ (i.e. strings having $3A's$ together which are redundant strings for above case(1)) $$=\frac{5!}{2!}$$
Hence, the total number of linear arrangements having $A's$ separated, is
$$\frac{7!}{3!2!}-\left(\frac{6!}{2!}-\frac{5!}{2!}\right)$$
$$=420-(360-60)=\color{blue}{120}$$
A: If you have $\_B\_N\_N\_S\_$ you can allocate three $A$'s among any of those $5$ empty spaces. That's $\binom{5}{3} = 10$ ways to allocate the $A$'s.
You then multiply that by the number of ways to arrange $B, N, N, S$ amongst themselves, which you have already done: $\frac{4!}{2!} = 12$.
All in all, that's $\binom{5}{3} \cdot \frac{4!}{2!} = 10 \cdot 12 = 120$ ways to arrange $BANANAS$ with all the $A$'s separated.
