Number of strings of length 8 using ABCDEF that contain ABC?












4












$begingroup$


Here is my attempt to calculate the number of strings of length $8$ using $6$ characters (ABCDEF) which contain 'ABC'. Basically I wanted to see if my approach is correct or if there is a better way to calculate this.



Basically we consider ABC like an element by itself, and it can go:



ABC X X X X X



X ABC X X X X



X X ABC X X X



X X X ABC X X



X X X X ABC X



X X X X X ABC



Where the X represents another character. So we have $6$ rows and in each row there are $6^5$ possibilities (can choose between $6$ chars. and have $5$ spots), therefore we have $6^6$ combinations.



However there are repetitions, which are the following cases:



ABC ABC X X



ABC X ABC X



ABC X X ABC



X ABC ABC X



X ABC X ABC



X X ABC ABC



So we subtract these possibilities which are $6^3$ ($6 cdot 6^2$ because there are $6$ rows and $2$ spots for $6$ chars in each row)



So we end up with $6^6 - 6^3$.



Thank you in advance.






combinatorics







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$endgroup$












  • $begingroup$
    Yeah good try. It seems to be the correct working to me!
    $endgroup$
    – Icycarus
    Dec 13 '18 at 23:37










  • $begingroup$
    Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 13 '18 at 23:40










  • $begingroup$
    Thanks for the quick responses, and the link, really needed it :)
    $endgroup$
    – m5signorini
    Dec 13 '18 at 23:43
















4












$begingroup$


Here is my attempt to calculate the number of strings of length $8$ using $6$ characters (ABCDEF) which contain 'ABC'. Basically I wanted to see if my approach is correct or if there is a better way to calculate this.



Basically we consider ABC like an element by itself, and it can go:



ABC X X X X X



X ABC X X X X



X X ABC X X X



X X X ABC X X



X X X X ABC X



X X X X X ABC



Where the X represents another character. So we have $6$ rows and in each row there are $6^5$ possibilities (can choose between $6$ chars. and have $5$ spots), therefore we have $6^6$ combinations.



However there are repetitions, which are the following cases:



ABC ABC X X



ABC X ABC X



ABC X X ABC



X ABC ABC X



X ABC X ABC



X X ABC ABC



So we subtract these possibilities which are $6^3$ ($6 cdot 6^2$ because there are $6$ rows and $2$ spots for $6$ chars in each row)



So we end up with $6^6 - 6^3$.



Thank you in advance.






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yeah good try. It seems to be the correct working to me!
    $endgroup$
    – Icycarus
    Dec 13 '18 at 23:37










  • $begingroup$
    Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 13 '18 at 23:40










  • $begingroup$
    Thanks for the quick responses, and the link, really needed it :)
    $endgroup$
    – m5signorini
    Dec 13 '18 at 23:43














4












4








4





$begingroup$


Here is my attempt to calculate the number of strings of length $8$ using $6$ characters (ABCDEF) which contain 'ABC'. Basically I wanted to see if my approach is correct or if there is a better way to calculate this.



Basically we consider ABC like an element by itself, and it can go:



ABC X X X X X



X ABC X X X X



X X ABC X X X



X X X ABC X X



X X X X ABC X



X X X X X ABC



Where the X represents another character. So we have $6$ rows and in each row there are $6^5$ possibilities (can choose between $6$ chars. and have $5$ spots), therefore we have $6^6$ combinations.



However there are repetitions, which are the following cases:



ABC ABC X X



ABC X ABC X



ABC X X ABC



X ABC ABC X



X ABC X ABC



X X ABC ABC



So we subtract these possibilities which are $6^3$ ($6 cdot 6^2$ because there are $6$ rows and $2$ spots for $6$ chars in each row)



So we end up with $6^6 - 6^3$.



Thank you in advance.






combinatorics







share|cite|improve this question











$endgroup$




Here is my attempt to calculate the number of strings of length $8$ using $6$ characters (ABCDEF) which contain 'ABC'. Basically I wanted to see if my approach is correct or if there is a better way to calculate this.



Basically we consider ABC like an element by itself, and it can go:



ABC X X X X X



X ABC X X X X



X X ABC X X X



X X X ABC X X



X X X X ABC X



X X X X X ABC



Where the X represents another character. So we have $6$ rows and in each row there are $6^5$ possibilities (can choose between $6$ chars. and have $5$ spots), therefore we have $6^6$ combinations.



However there are repetitions, which are the following cases:



ABC ABC X X



ABC X ABC X



ABC X X ABC



X ABC ABC X



X ABC X ABC



X X ABC ABC



So we subtract these possibilities which are $6^3$ ($6 cdot 6^2$ because there are $6$ rows and $2$ spots for $6$ chars in each row)



So we end up with $6^6 - 6^3$.



Thank you in advance.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 23:39









N. F. Taussig

44.5k103357




44.5k103357










asked Dec 13 '18 at 23:31









m5signorinim5signorini

233




233












  • $begingroup$
    Yeah good try. It seems to be the correct working to me!
    $endgroup$
    – Icycarus
    Dec 13 '18 at 23:37










  • $begingroup$
    Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 13 '18 at 23:40










  • $begingroup$
    Thanks for the quick responses, and the link, really needed it :)
    $endgroup$
    – m5signorini
    Dec 13 '18 at 23:43


















  • $begingroup$
    Yeah good try. It seems to be the correct working to me!
    $endgroup$
    – Icycarus
    Dec 13 '18 at 23:37










  • $begingroup$
    Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Dec 13 '18 at 23:40










  • $begingroup$
    Thanks for the quick responses, and the link, really needed it :)
    $endgroup$
    – m5signorini
    Dec 13 '18 at 23:43
















$begingroup$
Yeah good try. It seems to be the correct working to me!
$endgroup$
– Icycarus
Dec 13 '18 at 23:37




$begingroup$
Yeah good try. It seems to be the correct working to me!
$endgroup$
– Icycarus
Dec 13 '18 at 23:37












$begingroup$
Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 13 '18 at 23:40




$begingroup$
Welcome to MathSE. Your solution is correct. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 13 '18 at 23:40












$begingroup$
Thanks for the quick responses, and the link, really needed it :)
$endgroup$
– m5signorini
Dec 13 '18 at 23:43




$begingroup$
Thanks for the quick responses, and the link, really needed it :)
$endgroup$
– m5signorini
Dec 13 '18 at 23:43










1 Answer
1






active

oldest

votes


















2












$begingroup$

Your outcome is correct. This bruteforce calculation confirms it :D:



var count = 0;

for (let a0=0; a0<6; a0++)

for (let a1=0; a1<6; a1++)

for (let a2=0; a2<6; a2++)

for (let a3=0; a3<6; a3++)

for (let a4=0; a4<6; a4++)

for (let a5=0; a5<6; a5++)

for (let a6=0; a6<6; a6++)

for (let a7=0; a7<6; a7++) {

    let s =  a0+""+a1+""+a2+""+a3+""+a4+""+a5+""+a6+""+a7;

    if (s.indexOf("012")>=0) count++;

}

console.log(count);



46440




Another way to do this is to consider the state machine that tests whether a string contains the subsequence "ABC":



State machine for accepting words containing ABC



We can use this to calculate the probability of a 8 letter word containing the subsequence "ABC". It is equivalent to whether we reach the state "ABC" starting from the empty one in 8 steps. The matrix is



$$left [
begin{array}\
frac{5}{6} & frac{1}{6} & 0 & 0 \
frac{4}{6} & frac{1}{6} & frac{1}{6} & 0 \
frac{4}{6} & frac{1}{6} & 0 & frac{1}{6} \
0 & 0 & 0 & 1 \
end{array}
right ]
$$



Its eight power's top-right element gives the answer



(1/6[[5, 1, 0, 0], [4, 1, 1,0], [4, 1, 0, 1], [0,0,0,6]])^8


$$frac{46440}{6^8}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Cool! Thanks for the effort
    $endgroup$
    – m5signorini
    Dec 14 '18 at 14:04











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Your outcome is correct. This bruteforce calculation confirms it :D:



var count = 0;

for (let a0=0; a0<6; a0++)

for (let a1=0; a1<6; a1++)

for (let a2=0; a2<6; a2++)

for (let a3=0; a3<6; a3++)

for (let a4=0; a4<6; a4++)

for (let a5=0; a5<6; a5++)

for (let a6=0; a6<6; a6++)

for (let a7=0; a7<6; a7++) {

    let s =  a0+""+a1+""+a2+""+a3+""+a4+""+a5+""+a6+""+a7;

    if (s.indexOf("012")>=0) count++;

}

console.log(count);



46440




Another way to do this is to consider the state machine that tests whether a string contains the subsequence "ABC":



State machine for accepting words containing ABC



We can use this to calculate the probability of a 8 letter word containing the subsequence "ABC". It is equivalent to whether we reach the state "ABC" starting from the empty one in 8 steps. The matrix is



$$left [
begin{array}\
frac{5}{6} & frac{1}{6} & 0 & 0 \
frac{4}{6} & frac{1}{6} & frac{1}{6} & 0 \
frac{4}{6} & frac{1}{6} & 0 & frac{1}{6} \
0 & 0 & 0 & 1 \
end{array}
right ]
$$



Its eight power's top-right element gives the answer



(1/6[[5, 1, 0, 0], [4, 1, 1,0], [4, 1, 0, 1], [0,0,0,6]])^8


$$frac{46440}{6^8}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Cool! Thanks for the effort
    $endgroup$
    – m5signorini
    Dec 14 '18 at 14:04
















2












$begingroup$

Your outcome is correct. This bruteforce calculation confirms it :D:



var count = 0;

for (let a0=0; a0<6; a0++)

for (let a1=0; a1<6; a1++)

for (let a2=0; a2<6; a2++)

for (let a3=0; a3<6; a3++)

for (let a4=0; a4<6; a4++)

for (let a5=0; a5<6; a5++)

for (let a6=0; a6<6; a6++)

for (let a7=0; a7<6; a7++) {

    let s =  a0+""+a1+""+a2+""+a3+""+a4+""+a5+""+a6+""+a7;

    if (s.indexOf("012")>=0) count++;

}

console.log(count);



46440




Another way to do this is to consider the state machine that tests whether a string contains the subsequence "ABC":



State machine for accepting words containing ABC



We can use this to calculate the probability of a 8 letter word containing the subsequence "ABC". It is equivalent to whether we reach the state "ABC" starting from the empty one in 8 steps. The matrix is



$$left [
begin{array}\
frac{5}{6} & frac{1}{6} & 0 & 0 \
frac{4}{6} & frac{1}{6} & frac{1}{6} & 0 \
frac{4}{6} & frac{1}{6} & 0 & frac{1}{6} \
0 & 0 & 0 & 1 \
end{array}
right ]
$$



Its eight power's top-right element gives the answer



(1/6[[5, 1, 0, 0], [4, 1, 1,0], [4, 1, 0, 1], [0,0,0,6]])^8


$$frac{46440}{6^8}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Cool! Thanks for the effort
    $endgroup$
    – m5signorini
    Dec 14 '18 at 14:04














2












2








2





$begingroup$

Your outcome is correct. This bruteforce calculation confirms it :D:



var count = 0;

for (let a0=0; a0<6; a0++)

for (let a1=0; a1<6; a1++)

for (let a2=0; a2<6; a2++)

for (let a3=0; a3<6; a3++)

for (let a4=0; a4<6; a4++)

for (let a5=0; a5<6; a5++)

for (let a6=0; a6<6; a6++)

for (let a7=0; a7<6; a7++) {

    let s =  a0+""+a1+""+a2+""+a3+""+a4+""+a5+""+a6+""+a7;

    if (s.indexOf("012")>=0) count++;

}

console.log(count);



46440




Another way to do this is to consider the state machine that tests whether a string contains the subsequence "ABC":



State machine for accepting words containing ABC



We can use this to calculate the probability of a 8 letter word containing the subsequence "ABC". It is equivalent to whether we reach the state "ABC" starting from the empty one in 8 steps. The matrix is



$$left [
begin{array}\
frac{5}{6} & frac{1}{6} & 0 & 0 \
frac{4}{6} & frac{1}{6} & frac{1}{6} & 0 \
frac{4}{6} & frac{1}{6} & 0 & frac{1}{6} \
0 & 0 & 0 & 1 \
end{array}
right ]
$$



Its eight power's top-right element gives the answer



(1/6[[5, 1, 0, 0], [4, 1, 1,0], [4, 1, 0, 1], [0,0,0,6]])^8


$$frac{46440}{6^8}$$






share|cite|improve this answer









$endgroup$



Your outcome is correct. This bruteforce calculation confirms it :D:



var count = 0;

for (let a0=0; a0<6; a0++)

for (let a1=0; a1<6; a1++)

for (let a2=0; a2<6; a2++)

for (let a3=0; a3<6; a3++)

for (let a4=0; a4<6; a4++)

for (let a5=0; a5<6; a5++)

for (let a6=0; a6<6; a6++)

for (let a7=0; a7<6; a7++) {

    let s =  a0+""+a1+""+a2+""+a3+""+a4+""+a5+""+a6+""+a7;

    if (s.indexOf("012")>=0) count++;

}

console.log(count);



46440




Another way to do this is to consider the state machine that tests whether a string contains the subsequence "ABC":



State machine for accepting words containing ABC



We can use this to calculate the probability of a 8 letter word containing the subsequence "ABC". It is equivalent to whether we reach the state "ABC" starting from the empty one in 8 steps. The matrix is



$$left [
begin{array}\
frac{5}{6} & frac{1}{6} & 0 & 0 \
frac{4}{6} & frac{1}{6} & frac{1}{6} & 0 \
frac{4}{6} & frac{1}{6} & 0 & frac{1}{6} \
0 & 0 & 0 & 1 \
end{array}
right ]
$$



Its eight power's top-right element gives the answer



(1/6[[5, 1, 0, 0], [4, 1, 1,0], [4, 1, 0, 1], [0,0,0,6]])^8


$$frac{46440}{6^8}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 12:28









ploosu2ploosu2

4,6431024




4,6431024












  • $begingroup$
    Cool! Thanks for the effort
    $endgroup$
    – m5signorini
    Dec 14 '18 at 14:04


















  • $begingroup$
    Cool! Thanks for the effort
    $endgroup$
    – m5signorini
    Dec 14 '18 at 14:04
















$begingroup$
Cool! Thanks for the effort
$endgroup$
– m5signorini
Dec 14 '18 at 14:04




$begingroup$
Cool! Thanks for the effort
$endgroup$
– m5signorini
Dec 14 '18 at 14:04


















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