# 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?

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.

• I think that the OP's question is that 3A's should be separated.
– Koro
Jun 23, 2020 at 0:16
• Times the different arrangements of BNNS of course. Jun 23, 2020 at 11:16
• @user253751 Correct. This answer should be clarified to include that detail. Jun 23, 2020 at 11:41
• @rubik edited to include those details Jun 23, 2020 at 14:21

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=$$420-60=360$$

• Also this answer is good....for the BANANAS 😊. Jun 28, 2020 at 11:25

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