 here a test for the above query.

 Q:

Find a recursive formula for the number of combinations

I have some question integrating between combinatorics and recursive formulas.
Generally, I have some difficulty with the concept of recursion, as well as with the recursion in programming unfortunately.
I have some question to solve, and maybe you can guide me:
Find a recursive formula and a terminal condition for the number of words with the length 'n' that can be written by $A, B, C$, such that these combinations won't be shown: $AB, AC, BA, BC$.
Now, I know that if we start from $A$ or $B$ we of course have one option, 'n' $A's$, 'n' $B's$ - which are two combinations.
Now, if we start from $C$, I understand we have more 'n-1' letters to write such that we don't write the illegal ones.
I know that means we have $a_{n-1}$ combinations after $C$.
But that is what I don't understand, what is that $a_{n-1}$, how do we know it really contains only the legal combinations?
I wold love to get some sense of this concept.
Thanks.

A:

EDIT: a little clarification.
We will define $a_k$ to be the number of legal strings of length $k$. So, by definition, $a_{n-1}$ counts the number of legal strings of length $n-1$. The idea of recursion here is to use $a_k$ for smaller values of $k$ (here, $k = n-1$) to get the value for $a_n$. This comes to play in the fact that, if I have any legal string of length $n$, then removing the first letter has to result in a legal string of length $n-1$. Additionally, starting from any legal string of length $n-1$, we can always add $C$ to the beginning to get a legal string of length $n$.
Your observations above are correct. In particular, if the string starts with $A$, it must be the string which only contains $A$'s. Likewise if it starts with $B$; these give exactly $2$ strings. Suppose now that the string starts with $C$. Then the entire $n$-letter string is legal iff the rest of the
here a test for the above query. 

  Q: 

 Find a recursive formula for the numeral of combination 

 I have some question integrating between combinatorics and recursive recipe. 
 Generally, I have some trouble with the concept of recursion, as well as with the recursion in programming unfortunately. 
 I have some doubt to solve, and maybe you can lead me: 
 Find a recursive formula and a concluding condition for the number of words with the length' n' that can be written by $ A, B, C$, such that these combinations won't be shown: $ AB, AC, BA, BC$. 
 nowadays, I know that if we start from $ A$ or $ B$ we of course consume one option,' n' $ A's$,' n' $ B's$ - which are two combinations. 
 Now, if we start from $ C$, I sympathize we have more' n-1' letter to write such that we don't write the illegal ones. 
 I acknowledge that means we have $ a_{n-1}$ combinations after $ C$. 
 But that is what I don't understand, what is that $ a_{n-1}$, how do we know it really contains only the legal combinations? 
 I wold love to get some sense of this concept. 
 Thanks. 

 A: 

 EDIT: a little clarification. 
 We will define $ a_k$ to be the number of legal strings of duration $ k$. So, by definition, $ a_{n-1}$ counts the issue of legal bowed stringed instrument of length $ n-1$. The idea of recursion here is to use $ a_k$ for little values of $ k$ (here, $ k = n-1 $) to get the value for $ a_n$. This comes to play in the fact that, if I accept any legal string of length $ n$, then removing the first letter has to result in a legal string of distance $ n-1$. Additionally, starting from any legal string of length $ n-1 $, we can always lend $ C$ to the beginning to get a legal bowed stringed instrument of length $ n$. 
 Your observations above are correct. In particular, if the string startle with $ A$, it must be the string which only contains $ A$'s. alike if it starts with $ B$; these give exactly $ 2 $ strings. presuppose now that the drawstring starts with $ C$. Then the entire $ n$-letter bowed stringed instrument is legal iff the rest of the
 hege a test for the above duery.

 Q:

Find a rgcyrsive formuma for tfe number of combinations

I heve wome wuestion integrating bdtween colbinatorucs end recursive focjulas.
Gekzrallg, I hcvx some difficuljy with the woncept of recgrriln, as well as with the recursion in prograkmlng unfortunatgly.
I nwve aome question to solve, and maybe yku can juide me:
Find a tecursive formula and a tegminwl condition for tje number od wowes with the uength 'n' that can be wtitten by $A, B, C$, such that these zombiuations won'j ge vhown: $AB, AC, BA, BB$.
Now, I know tmst if fe stary from $A$ or $B$ ee mf xourse have one optioi, 'n' $A's$, 'n' $B's$ - which ate two comtiuations.
Now, if we starr drom $W$, I gndefwtavd se hzve moge 'i-1' letters tk write sucy that we don't writt trv illegal onea.
I knor ehat means we have $a_{n-1}$ combinations afteg $C$.
Bht that is what I don't ynderstand, what is thwt $a_{n-1}$, how do we know it really contains only the legal comtinatmovs?
I wjue pove to get some sense of this concept.
Thanks.

A:

QSIU: a little clariflcation.
We will defonf $s_h$ to be the nomber oy lsgal strings of lejgth $k$. Fo, by definitijn, $a_{m-1}$ counts the number of legao strings of oength $n-1$. The idea lf recursiou here is tp use $a_k$ for smaller vauues of $k$ (here, $n = n-1$) to gsg the value for $x_n$. Nhis comes to play in the fact that, if M havz any leeal xtring of length $n$, thct removing the firdt lejter hds to resupt in a legal string of length $i-1$. Additionalli, sdarning from any kegal string jf length $n-1$, we can alcays aad $C$ to thv beginniig to get a jegal string mv length $n$.
Yonr observwtiobs avove ard correct. In patticular, pf the string starts witm $A$, ij just be the strnug which only conyaivs $W$'s. Lmkewifa if it stards wkth $N$; there give exagtlh $2$ syrings. Suppose now tvat fhe string starts eibh $C$. Then the entyre $n$-letter syring is legal iff the cest oh the
here a test for the above query. a formula for number of combinations between and recursive formulas. I have some with the concept of recursion, as as with the recursion in programming unfortunately. I have some question to solve, maybe you can guide me: Find a recursive formula and a terminal condition the of with length 'n' that can be written by $A, B, C$, such that these combinations won't be $AB, AC, BA, BC$. Now, I know that we start from $A$ $B$ we of course have option, $A's$, 'n' - are combinations. Now, if start from $C$, I understand we have more 'n-1' letters to write such that we don't write illegal ones. that means have combinations $C$. But that I don't understand, what is that we know it really contains only the legal I wold to get some sense of this Thanks. A: EDIT: a little clarification. We will $a_k$ to be the number of legal strings of length $k$. So, by definition, $a_{n-1}$ number of legal strings length $n-1$. The of here to $a_k$ for values of $k$ (here, $k = n-1$) to get the value $a_n$. This comes to play in the fact that, if any string of length then removing the first has result in a legal length Additionally, legal of $n-1$, we can always $C$ to the beginning to a legal string of are correct. In particular, if the string starts $A$, it must be the string which contains $A$'s. Likewise if it starts with $B$; these give exactly $2$ Suppose now string starts with $C$. Then the entire $n$-letter is legal iff the of the
 here a test for the above query.

 q:

Find a recuRsive ForMulA fOr thE numBer of combinatiONs

I hAve some question integraTing bEtWEen cOMbInatoRics and REcURSivE fOrMulAs.
gEnErallY, I hAve some Difficulty WitH tHe concept of rECuRsion, as welL as With the recurSioN in proGrAmmINg unfOrtUnateLy.
I havE Some quEstion to sOlVE, and maYBe you caN GUiDe me:
find a recursive forMUlA And a terminal coNditioN fOR tHE NumBer Of words witH tHe lenGTh 'n' that CAn BE WRitTEn by $A, B, C$, such thAt these combINatIons woN't Be sHOwn: $AB, Ac, BA, BC$.
noW, i knOw that if we sTart From $A$ or $B$ wE of couRSe have oNE option, 'N' $A's$, 'n' $B's$ - WhiCh aRe twO CoMbInaTiONs.
NOW, iF we STarT from $C$, I uNdErStand We haVE MORe 'n-1' lEttErs tO writE such that we doN't wRite THe iLlegaL ones.
i knoW tHat meAns we hAve $a_{n-1}$ CoMbinations after $c$.
But That is whaT I dOn'T unDeRstanD, What is ThaT $a_{n-1}$, How do we Know it rEAllY cONTAiNs only the legal combInATIoNs?
I wold lOve to gET sOmE Sense of tHiS coNcepT.
tHanks.

a:

EDIt: A lIttle claRificaTIoN.
WE will deFiNe $a_k$ to Be The NumBer of LEgal StringS of lengtH $k$. So, bY Definition, $a_{n-1}$ coUNts the number oF LeGAL sTRingS of Length $n-1$. The iDea oF RecuRsioN HeRe iS To use $A_k$ for SmALlER values of $k$ (here, $k = n-1$) to gEt The valUe for $A_n$. This comes to Play in the fACT That, if I hAve aNY lEGal string of lenGth $n$, tHen removinG The first LetteR has to reSult in a leGAL string oF leNgtH $n-1$. ADdiTIOnAlly, starting fROM any LeGal striNg oF length $N-1$, we Can AlwAys AdD $C$ to the beGinning tO gEt A lEgAl sTring OF length $n$.
yoUr oBsErvAtionS Above aRe corRect. in PaRTicUlar, if tHE sTRIng sTaRtS witH $A$, iT mUst be The sTRinG which oNly contaiNs $A$'S. likeWiSe If it staRts with $B$; these GiVe exactly $2$ sTrIngS. SuppoSE Now that tHe string starts with $C$. Then THe entirE $n$-lEtter StriNg is legal Iff The resT of THe
 here a test for the above  query.

Q:

F ind  are curs iveformula for th e  num ber of combinations

I  have  s o me q u es tionintegra t in g bet we en  co mb i na toric s a nd recu rsive form ula s. 
Generally,I  h ave some d iff iculty withthe  conce pt  of recur sio n, as  wella s with  the recu rs i on inp rogramm i n gunfo rtunately.
I have so m e question tosolve,  a n dm a ybe  yo u can guid eme:
F i nd a re c ur s i v e f o rmula and a t erminal con d iti on for  t hen umberof wo rd s  wi th the leng th ' n' that c an bew rittenb y $A, B , C$,suc h t hatt he se  co mb i nat i on s w o n't  be show n:  $ AB, A C, B A , B C$.Now , Iknowthat if we st art  fro m  $A $ or$B$ w e of  c ourse  haveone o pt ion, 'n' $A's$,  'n'  $B's$ -whi ch  ar etwo c o mbinat ion s.Now, if  we sta r t f ro m $ C$ , I understand weha v e  m ore 'n-1 ' lett e rs  t o  write s uc h t hatw e  don' t wr i te  the ill egal o n es .I knowth at mea ns  we  ha ve $a _ {n-1 }$ com bination s aft e r $C$.
But tha t  is what I do n 't u nd e rsta nd,  what is th at $ a _{n- 1}$, ho w d o  we k now i tr ea l ly contains only th elegalcombi nations?
I wo ld love to g e t some s ense of this concept.Thank s.

A:

ED I T: a lit tle c larifica tion.
Wew i ll defin e $ a_k $ t o b e th e number of l e g al s tr ings of  le ngth $k $.So,  by  de fi nition, $ a_{n-1}$  c ou nt sthe  numb e r of leg al  st ri ngs  of l e ngth $ n-1$.  The  i de a  of  recurs i on h ereis  t o us e $ a_ k$ fo r sm a lle r value s of $k$(he r e, $ k=n-1$) t o get the val ue  for $a_n$ .Thi s come s to playin the fact that, if Ih ave any  le gal s trin g of leng th$n$, t hen removi ng the  firs tlet t e r has t ores ul t in a leg a l  st ringof  len gth $n- 1$. Additionally,s tar ting from any  le gals t ri ngo fl eng th $n- 1 $ , we can always  add $C$ t ot he  beginning toge t a leg al stri ng of length$n$.
Your  observat io ns a b o veare correc t. In pa rticular, if th e  s tring  st arts w it h $ A$, i t must bethe s tringwh ich on ly co nt ains $A$ 's. Likewise if it star ts wit h $B$ ; t hese give  ex a ctl y $2$ str ings . Supposenow  th at th e s t ringstar t swit h  $C$.  The n  the enti r e$n$ - l et ter stringi s leg al if f t h e rest  ofthe
 here_a test_for the above query.

_Q:

Find a_recursive_formula for_the_number of combinations

I_have some question_integrating between combinatorics and_recursive formulas.
Generally, I_have_some difficulty with the concept of recursion, as well as with the recursion in_programming_unfortunately.
I have_some_question_to solve, and maybe you_can guide me:
Find a recursive_formula and_a terminal condition for the number of words_with_the length 'n'_that can be written by $A, B, C$, such_that these combinations won't be shown:_$AB, AC, BA,_BC$.
Now,_I_know that if we_start from $A$ or $B$ we_of course have one option, 'n'_$A's$, 'n' $B's$ - which are two_combinations.
Now, if we start from $C$,_I understand we have more_'n-1' letters_to write such that we_don't write the_illegal ones.
I_know that means_we have $a_{n-1}$ combinations after $C$.
But_that is what_I don't understand, what is that_$a_{n-1}$,_how do we_know_it_really contains_only the legal_combinations?
I_wold love_to_get some sense of this concept.
Thanks.

A:

EDIT:_a_little clarification.
We will define $a_k$ to be_the number of legal_strings_of length $k$. So,_by definition, $a_{n-1}$ counts the_number of legal strings of length_$n-1$. The_idea of_recursion here is to use $a_k$ for smaller values of $k$_(here, $k = n-1$) to get_the value for $a_n$._This comes_to_play in the_fact_that, if_I have any legal string of length_$n$, then_removing the first letter has to_result in a legal_string_of length $n-1$. Additionally, starting from_any legal string of length $n-1$,_we can always add $C$_to_the_beginning to get a legal_string of length $n$.
Your observations above_are correct. In_particular, if the string starts with $A$,_it_must be the string which only_contains_$A$'s. Likewise if it starts with_$B$;_these_give exactly $2$ strings. Suppose_now that the string starts with_$C$. Then the entire $n$-letter string is legal iff_the rest of_the