Mixing
How many word can we make with infinited times of $B$, $D$ $M$ and only one $O$?
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In a case, we have infinite times of the letter B, D, M and only one O. How many different word containing those letter can we make (can be meaningless in this term) ?
combinatorics combinatorics-on-words
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
$endgroup$
add a comment |
$begingroup$
In a case, we have infinite times of the letter B, D, M and only one O. How many different word containing those letter can we make (can be meaningless in this term) ?
combinatorics combinatorics-on-words
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
$endgroup$
2
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
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– Exp ikx
Jan 19 at 12:46
1
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
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Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45
add a comment |
$begingroup$
In a case, we have infinite times of the letter B, D, M and only one O. How many different word containing those letter can we make (can be meaningless in this term) ?
combinatorics combinatorics-on-words
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
$endgroup$
In a case, we have infinite times of the letter B, D, M and only one O. How many different word containing those letter can we make (can be meaningless in this term) ?
combinatorics combinatorics-on-words
combinatorics combinatorics-on-words
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
edited Feb 4 at 3:53
Anirban Niloy
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
asked Jan 19 at 12:39
Anirban NiloyAnirban Niloy
606218
606218
2
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
$endgroup$
– Exp ikx
Jan 19 at 12:46
1
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
$begingroup$
Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45
add a comment |
2
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
$endgroup$
– Exp ikx
Jan 19 at 12:46
1
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
$begingroup$
Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45
2
2
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
$endgroup$
– Exp ikx
Jan 19 at 12:46
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
$endgroup$
– Exp ikx
Jan 19 at 12:46
1
1
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
$begingroup$
Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you have infinite letters of at least 2 types you can make infinite words.
Example:
B, BD, BBD, BBBD, ...
If you mean words with length n:
Denote the ways to make a word with infinite B,D,M and length k by G(k).Then the answer you need is G(n) + n * G(n-1).
To find G(n), let's say the word is _ _ _ ... _ _ _. Then each blank can be B, D or M so G(n) = 3^n
Therefore You can make 3^n + n * 3^(n-1) words with length n with infinite B, D and M and 1 O
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
$endgroup$
add a comment |
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$begingroup$
As you have infinite letters of at least 2 types you can make infinite words.
Example:
B, BD, BBD, BBBD, ...
If you mean words with length n:
Denote the ways to make a word with infinite B,D,M and length k by G(k).Then the answer you need is G(n) + n * G(n-1).
To find G(n), let's say the word is _ _ _ ... _ _ _. Then each blank can be B, D or M so G(n) = 3^n
Therefore You can make 3^n + n * 3^(n-1) words with length n with infinite B, D and M and 1 O
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
$endgroup$
add a comment |
$begingroup$
As you have infinite letters of at least 2 types you can make infinite words.
Example:
B, BD, BBD, BBBD, ...
If you mean words with length n:
Denote the ways to make a word with infinite B,D,M and length k by G(k).Then the answer you need is G(n) + n * G(n-1).
To find G(n), let's say the word is _ _ _ ... _ _ _. Then each blank can be B, D or M so G(n) = 3^n
Therefore You can make 3^n + n * 3^(n-1) words with length n with infinite B, D and M and 1 O
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
$endgroup$
add a comment |
$begingroup$
As you have infinite letters of at least 2 types you can make infinite words.
Example:
B, BD, BBD, BBBD, ...
If you mean words with length n:
Denote the ways to make a word with infinite B,D,M and length k by G(k).Then the answer you need is G(n) + n * G(n-1).
To find G(n), let's say the word is _ _ _ ... _ _ _. Then each blank can be B, D or M so G(n) = 3^n
Therefore You can make 3^n + n * 3^(n-1) words with length n with infinite B, D and M and 1 O
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
$endgroup$
As you have infinite letters of at least 2 types you can make infinite words.
Example:
B, BD, BBD, BBBD, ...
If you mean words with length n:
Denote the ways to make a word with infinite B,D,M and length k by G(k).Then the answer you need is G(n) + n * G(n-1).
To find G(n), let's say the word is _ _ _ ... _ _ _. Then each blank can be B, D or M so G(n) = 3^n
Therefore You can make 3^n + n * 3^(n-1) words with length n with infinite B, D and M and 1 O
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
answered Jan 19 at 12:46
Gareth MaGareth Ma
526215
526215
add a comment |
add a comment |
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2
$begingroup$
If the words can be meaningless and infinite copies of letters are available, we can make infinitely many strings of letters and call them words.
$endgroup$
– Exp ikx
Jan 19 at 12:46
1
$begingroup$
If you mean making words with length k, I answered
$endgroup$
– Gareth Ma
Jan 19 at 12:46
$begingroup$
Sorry. It is my fault. The word should be consisted of only 4 letters and that can be constructed in many ways and we can surely approach with the permutation process. No matter whatever the word means. I beg your pardon and I didn't understand the length of the word 'k'.
$endgroup$
– Anirban Niloy
Jan 19 at 12:51
$begingroup$
@AnirbanNiloy Oh... k is just any variable :D no problem man gl
$endgroup$
– Gareth Ma
Jan 19 at 14:45