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I found a problem in my counting textbook, which is stated above. It gives the string AAAAAABBBBBB, and asks for how many arrangements (using all of the letters) there are such that every A is next to at least one other A.

I calculated and then looked into the back for the answer, and the answer appears to be $105$. My answer fell short of that by quite a bit. I broke down the string into various cases, and then used Stars and Bars to find how many possibilities there are for each. Now here's what I have got so far. First case would be all As are right next to each other, leaving $2$ spots for the $6$ Bs. That gives $\binom{7}{1}$ from Stars and Bars, $7$ possibilities. Second case was dividing the As into $2$ groups of 3 As. There would have to be $1$ B between the two, which leaves $5$ Bs that can be moved. Using Stars and Bars, there are $3$ possible places to place a B and $5$ Bs in total, so $\binom{6}{2}$, $15$ possibilities. Then there's a group of $4$ As and another group of $2$ As. $1$ B would be placed inbetween, and then the calculation would be the same as the second case, except it would have to be doubled to account that the groups of As can be swapped and it would be distinct. That gives $30$ possibilities. Then I found one final case of dividing the As into 3 groups of 2 As. 2 Bs would immediately be placed between the 3 groups, leaving 4 Bs to move between the 4 possible locations. I got $\binom{5}{3}$ for that, which adds $10$ possibilities. Summing it up, I only have $62$ possibilities, which is quite far from the $105$ answer. Any ideas where I might have miscalculated or missed a potential case? Additionally, are there any better ways to calculate this compared to this method of casework?

RobPratt
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Boris Poris
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4 Answers4

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You missed the fact that there can be any number of Bs between the two groups of As. For the $3+3$ case,

If there's one B, number of cases is $6$.

If there's two Bs, number of cases is $5$.

If there's three Bs, number of cases is $4$. $$\cdots$$ If there's six Bs, number of cases is $1$.

In total, there are $6+5+4+3+2+1=21$ ways.

Similarly, there are $21\cdot2=42$ ways for the $4+2$ case.

Now, a similar mistake you did for the $2+2+2$ case. Can you get it now and complete your solution?

Hope this helps. Ask anything if not clear :)

ultralegend5385
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  • I thought I accounted for that while initially calculating. For example, in the 3+3 case, I split into 3 groups for the remaining Bs. First group goes before the first group of As, second goes after where the first B was immediately assigned, and the third goes behind the second group of As. But I have discovered another issue, which is I forgot to account for that the "bars" could be on top of each other (so the 1 B only scenario for example would have been skipped). Upon calculating with that, it appears we get the same amount. – Boris Poris Apr 08 '21 at 03:05
  • Thanks for helping me to realize that mistake! Upon calculating again (counting the scenarios where the "bars" can be on top of each other), it appears I now have the full $105$ possibilities. Also, I'm curious if this casework method be the best way to approach this problem, or do you know of any other methods that could be more efficient? – Boris Poris Apr 08 '21 at 03:09
  • @BorisPoris: Due to the restriction of every A next to some A, afaik there's no other efficient way. – ultralegend5385 Apr 08 '21 at 03:39
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If we space Bs first, there are 7 empty positions:

\begin{align}-B-B-B-B-B-B-\end{align}

There are 4 ways to insert A:

6A group together, $C_1^7$=7 ways

4A and 2A group together, 2$C_2^7$=42 ways

3A and 3A together,$C_2^7$ =21 ways

2A,2A,2A together, $C_3^7$ =35 ways

Total 105 ways.

Star Bright
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  • There would be $7$ possibilities, but what the case where there's one A to the side and the other 5 As in a group also would not work, since all As have to be next to at least one other A. – Boris Poris Apr 08 '21 at 02:35
  • "No A's are next to any other A's" is not the complement of "Each A is next to another A," so this does not work. – Mike Earnest Apr 08 '21 at 02:48
  • @ Boris, I updated my answer. Hope it is helpful. – Star Bright Apr 08 '21 at 03:35
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Another way which might work is to calculate non-favourable cases. every arrangement with BAB in it is not favourable(case 1).also,arrangements with starting AB(case2),arrangements ending with BA(case 3) are not favourable.but,while calculating case 2 and 3 we should avoid arrangements with BAB as they are repeated in case 1

IITM
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One way is using generating functions:

Once you space the B's, there are seven slots for placing A's.

Each slot can have either 0, or 2 or more A's. The "generating function" for a single slot is a polynomial $a_0+a_1x+a_2x^2+...$ where $a_k$ is the number of ways to put $k$ A's into a single slot. That is, the generating function is $f(x)=1+x^2+x^3+...=\frac{1}{1-x}-x=\frac{1-x+x^2}{1-x}$.

But you don't have 1 slot, you have 7. The generating function for 7 slots is $g(x)=f(x)^7=\frac{(1-x+x^2)^7}{(1-x)^7}=1+0x+7x^2+7x^3+28x^4+49x^5+105x^6+196x^7+37x^8...$.

Each term $b_kx^k$ gives the number of ways $b_k$ to place $k$ A's into the 7 slots, subject to your restriction. Since you have six A's, you need to find the coefficient of $x^6$ in the taylor series, which is 105.

Admittedly, the last step is messy on paper, since the 6th derivative of $g(x)$ might get complicated, but it's easy for a computer algebra package.

Michael Hartley
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