How would I go about finding how many combinations based on this criteria. [closed]












-1












$begingroup$


Alright, say I have a 30 letter string.




  • I can put . or + between any two letters

  • There can be any amount of . and + in a "modified string"

  • I cannot put . or + before the first letter or after the last letter

  • I cannot put . next to ., + next to + or . next to +, in any order.






combinatorics







share|cite|improve this question









$endgroup$



closed as off-topic by Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo Jan 10 at 9:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
    $endgroup$
    – lulu
    Jan 10 at 0:58
















-1












$begingroup$


Alright, say I have a 30 letter string.




  • I can put . or + between any two letters

  • There can be any amount of . and + in a "modified string"

  • I cannot put . or + before the first letter or after the last letter

  • I cannot put . next to ., + next to + or . next to +, in any order.






combinatorics







share|cite|improve this question









$endgroup$



closed as off-topic by Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo Jan 10 at 9:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
    $endgroup$
    – lulu
    Jan 10 at 0:58














-1












-1








-1





$begingroup$


Alright, say I have a 30 letter string.




  • I can put . or + between any two letters

  • There can be any amount of . and + in a "modified string"

  • I cannot put . or + before the first letter or after the last letter

  • I cannot put . next to ., + next to + or . next to +, in any order.






combinatorics







share|cite|improve this question









$endgroup$




Alright, say I have a 30 letter string.




  • I can put . or + between any two letters

  • There can be any amount of . and + in a "modified string"

  • I cannot put . or + before the first letter or after the last letter

  • I cannot put . next to ., + next to + or . next to +, in any order.






combinatorics




combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 0:56









dakedake

11




11




closed as off-topic by Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo Jan 10 at 9:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo Jan 10 at 9:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, max_zorn, mrtaurho, Abcd, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
    $endgroup$
    – lulu
    Jan 10 at 0:58


















  • $begingroup$
    What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
    $endgroup$
    – lulu
    Jan 10 at 0:58
















$begingroup$
What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
$endgroup$
– lulu
Jan 10 at 0:58




$begingroup$
What have you tried? Though I am not sure the rules are clear. What is the answer if you start with a $2$ letter string?
$endgroup$
– lulu
Jan 10 at 0:58










1 Answer
1






active

oldest

votes


















3












$begingroup$

Assuming that there are no . or + in the original string:



$3^{29} = 68630377364883$.



Each consecutive pair of letters will either have a ., a +, or neither in between them. These choices are independent, i.e. your choice for the first pair of letters does not effect what choices are available for the second. There are three choices for each pair (., +, or nothing) and 29 pairs, so you just multiply each independent choice: $3cdot3cdot3...=3^{29} = 68630377364883$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Are you sure it's not 29^3? That number seems incredibly high.
    $endgroup$
    – dake
    Jan 10 at 1:43




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Assuming that there are no . or + in the original string:



$3^{29} = 68630377364883$.



Each consecutive pair of letters will either have a ., a +, or neither in between them. These choices are independent, i.e. your choice for the first pair of letters does not effect what choices are available for the second. There are three choices for each pair (., +, or nothing) and 29 pairs, so you just multiply each independent choice: $3cdot3cdot3...=3^{29} = 68630377364883$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Are you sure it's not 29^3? That number seems incredibly high.
    $endgroup$
    – dake
    Jan 10 at 1:43


















3












$begingroup$

Assuming that there are no . or + in the original string:



$3^{29} = 68630377364883$.



Each consecutive pair of letters will either have a ., a +, or neither in between them. These choices are independent, i.e. your choice for the first pair of letters does not effect what choices are available for the second. There are three choices for each pair (., +, or nothing) and 29 pairs, so you just multiply each independent choice: $3cdot3cdot3...=3^{29} = 68630377364883$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Are you sure it's not 29^3? That number seems incredibly high.
    $endgroup$
    – dake
    Jan 10 at 1:43
















3












3








3





$begingroup$

Assuming that there are no . or + in the original string:



$3^{29} = 68630377364883$.



Each consecutive pair of letters will either have a ., a +, or neither in between them. These choices are independent, i.e. your choice for the first pair of letters does not effect what choices are available for the second. There are three choices for each pair (., +, or nothing) and 29 pairs, so you just multiply each independent choice: $3cdot3cdot3...=3^{29} = 68630377364883$






share|cite|improve this answer









$endgroup$



Assuming that there are no . or + in the original string:



$3^{29} = 68630377364883$.



Each consecutive pair of letters will either have a ., a +, or neither in between them. These choices are independent, i.e. your choice for the first pair of letters does not effect what choices are available for the second. There are three choices for each pair (., +, or nothing) and 29 pairs, so you just multiply each independent choice: $3cdot3cdot3...=3^{29} = 68630377364883$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 1:05









The Zach ManThe Zach Man

887




887












  • $begingroup$
    Are you sure it's not 29^3? That number seems incredibly high.
    $endgroup$
    – dake
    Jan 10 at 1:43




















  • $begingroup$
    Are you sure it's not 29^3? That number seems incredibly high.
    $endgroup$
    – dake
    Jan 10 at 1:43


















$begingroup$
Are you sure it's not 29^3? That number seems incredibly high.
$endgroup$
– dake
Jan 10 at 1:43






$begingroup$
Are you sure it's not 29^3? That number seems incredibly high.
$endgroup$
– dake
Jan 10 at 1:43





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