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Okay, I tried something, got close, but no idea why my answer is wrong.
I pick C >> $n-1$ options.
I Pick A (Here is problem):
A = n-1 options -(subtracting) (AB = n-2 options - ABB = n-3 options))
Which gets me:
$a_n = 2a_{n-1} - a_{n-2} + a_{n-3}$
Why is my answer wrong? I used the two mistakes.
Why is it: $a_n = 2a_{n-1} + a_{n-2}$

  • @lulu Oh I remember you, you helped me once hehe, thanks for before :) About this question, I see what you say,although yea, its almost the same. but anyway, I want to know why my solution is wrong, I used the problems. –  Aug 13 '22 at 11:44
  • Also, problem is I dont use series of an, bn, cn, I saw a video regarding this, but he used an bn and cn, whoever will think of using such stuff.. I use only an –  Aug 13 '22 at 11:46
  • As I understand your approach, you are prepending a character before a substring that is already without AA and BB. You have consider the strings with first character either A or C; what about those that start with B? – peterwhy Aug 13 '22 at 11:49
  • @peterwhy hmm, you mean to add the case of start with B? Ill update in a minute. –  Aug 13 '22 at 12:12
  • You wrote to subtract those "(AB = n-2 options - ABB = n-3 options)", which is to subtract the number of strings that have length $n$ and start with either ABA or ABC. May I understand why to subtract this? – peterwhy Aug 13 '22 at 12:19
  • @peterwhy because BB and AA are not good. and I saw a trick like this in other exercise. –  Aug 13 '22 at 12:22
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    Can't really follow your approach. I'd do it by conditioning on the starting term. Let $A_n,B_n, C_n$ denote good strings that begin with $A,B,C$. $T_n=A_n+B_n+C_n$ is the total. Then $C_n=T_{n-1}$, $A_n=B_{n-1}+C_{n-1}$ and $B_n=A_{n-1}+C_{n-1}$ and, of course $A_n=B_n$. That's enough. – lulu Aug 13 '22 at 12:23
  • @lulu Thanks, but I dont use mini-series for it, I dont do $a_n$ $b_n$ $c_n$ and such.. I use only $a_n$ –  Aug 13 '22 at 12:24
  • Sure, but as I say I really can't follow your reasoning. – lulu Aug 13 '22 at 12:25
  • @lulu I dont have time to study that thing, the test is tommorow :\ ( final exam ) –  Aug 13 '22 at 12:25
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    It's hard to just consider the start without conditioning, because a good string could begin $ABABAB\cdots$ or something like that. You don't know that you'll ever get to a $C$. But the conditioning makes it easy to handle. – lulu Aug 13 '22 at 12:26
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    Can't help you there. Conditioning on the start of a string is a standard counting method, it's not exotic. Good luck. – lulu Aug 13 '22 at 12:27
  • @lulu Thanks anyway :) I will try to see how to answer this question my way :) –  Aug 13 '22 at 12:27
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    I understand that AA and BB are not good; but you tried to remove ABA... and ABC... strings from A... strings, and that remains strings that start with either (3-substrings) AAB, AAC, ACA, ACB, or ACC. Some not-good strings are still here, and some good strings are gone. – peterwhy Aug 13 '22 at 12:30
  • @peterwhy ahh, I will try to see what to do, I dont know what to do, but I will try my hardest and see :\ Thanks –  Aug 13 '22 at 12:31
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    If you continue with @lulu 's approach, then $$\begin{align} T_n &= A_n + B_n + C_n\ &= \underbrace{(B_{n-1}+C_{n-1})+(A_{n-1}}{T{n-1}}+C_{n-1}) + T_{n-1}\ &= T_{n-1}+T_{n-2} + T_{n-1} \end{align}$$ – peterwhy Aug 13 '22 at 12:43
  • @peterwhy Thanks, but I will return to it later, I will try again after clearing my head from the exercise :) –  Aug 13 '22 at 12:51

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