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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?

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3 Answers 3

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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.

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    $\begingroup$ I think that the OP's question is that 3A's should be separated. $\endgroup$
    – Koro
    Jun 23, 2020 at 0:16
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    $\begingroup$ Times the different arrangements of BNNS of course. $\endgroup$
    – user253751
    Jun 23, 2020 at 11:16
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    $\begingroup$ @user253751 Correct. This answer should be clarified to include that detail. $\endgroup$
    – rubik
    Jun 23, 2020 at 11:41
  • $\begingroup$ @rubik edited to include those details $\endgroup$
    – user525966
    Jun 23, 2020 at 14:21
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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=$420-60=360$

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    $\begingroup$ Also this answer is good....for the BANANAS 😊. $\endgroup$
    – Sebastiano
    Jun 28, 2020 at 11:25
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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 $$\frac{7!}{3!2!}-\left(\frac{6!}{2!}-\frac{5!}{2!}\right)$$ $$=420-(360-60)=\color{blue}{120}$$

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