Arrangements of BANANAS where the A's are separated

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.


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=$420-60=360$


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 $$\frac{7!}{3!2!}-\left(\frac{6!}{2!}-\frac{5!}{2!}\right)$$ $$=420-(360-60)=\color{blue}{120}$$

Tags:

Combinatorics

Discrete Mathematics

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