How many ways can I arrange 5 As, 6 Bs, and 3 Cs [duplicate]
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This question already has an answer here:
Arranging 5 D's, 6 E's and 3 F's such that the First D precedes the First E which precedes the First F
2 answers
With requirement that $A$ precedes the first $B$ which precedes the first $C$.
Example: $AABABCBBAACBCB$ is correct and $BAAACBAACBCBBBB$ is incorrect.
I am thinking of using permutation, get the total number of ways of arranging them and divide the repeat.
There are $14$ spaces, $5$ ways of placing the first one ($5As$), $10$ ways of placing the second one $4As + 6Bs$, but how many ways to place $3rd$, $4th$ and the rest?
combinatorics permutations
asked Dec 2 '18 at 23:17
CupCup
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Dec 2 '18 at 23:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Arranging 5 D's, 6 E's and 3 F's such that the First D precedes the First E which precedes the First F
2 answers
With requirement that $A$ precedes the first $B$ which precedes the first $C$.
Example: $AABABCBBAACBCB$ is correct and $BAAACBAACBCBBBB$ is incorrect.
I am thinking of using permutation, get the total number of ways of arranging them and divide the repeat.
There are $14$ spaces, $5$ ways of placing the first one ($5As$), $10$ ways of placing the second one $4As + 6Bs$, but how many ways to place $3rd$, $4th$ and the rest?
combinatorics permutations
asked Dec 2 '18 at 23:17
CupCup
276
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marked as duplicate by N. F. Taussig combinatorics
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Dec 2 '18 at 23:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
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– lulu
Dec 2 '18 at 23:19
add a comment |
$begingroup$
This question already has an answer here:
Arranging 5 D's, 6 E's and 3 F's such that the First D precedes the First E which precedes the First F
2 answers
With requirement that $A$ precedes the first $B$ which precedes the first $C$.
Example: $AABABCBBAACBCB$ is correct and $BAAACBAACBCBBBB$ is incorrect.
I am thinking of using permutation, get the total number of ways of arranging them and divide the repeat.
There are $14$ spaces, $5$ ways of placing the first one ($5As$), $10$ ways of placing the second one $4As + 6Bs$, but how many ways to place $3rd$, $4th$ and the rest?
combinatorics permutations
asked Dec 2 '18 at 23:17
CupCup
276
$endgroup$
This question already has an answer here:
Arranging 5 D's, 6 E's and 3 F's such that the First D precedes the First E which precedes the First F
2 answers
With requirement that $A$ precedes the first $B$ which precedes the first $C$.
Example: $AABABCBBAACBCB$ is correct and $BAAACBAACBCBBBB$ is incorrect.
I am thinking of using permutation, get the total number of ways of arranging them and divide the repeat.
There are $14$ spaces, $5$ ways of placing the first one ($5As$), $10$ ways of placing the second one $4As + 6Bs$, but how many ways to place $3rd$, $4th$ and the rest?
This question already has an answer here:
Arranging 5 D's, 6 E's and 3 F's such that the First D precedes the First E which precedes the First F
2 answers
combinatorics permutations
combinatorics permutations
asked Dec 2 '18 at 23:17
CupCup
276
asked Dec 2 '18 at 23:17
CupCup
276
asked Dec 2 '18 at 23:17
CupCup
276
asked Dec 2 '18 at 23:17
CupCup
276
asked Dec 2 '18 at 23:17
CupCup
276
276
marked as duplicate by N. F. Taussig combinatorics
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Dec 2 '18 at 23:42
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Dec 2 '18 at 23:42
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
$endgroup$
– lulu
Dec 2 '18 at 23:19
add a comment |
$begingroup$
Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
$endgroup$
– lulu
Dec 2 '18 at 23:19
$begingroup$
Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
$endgroup$
– lulu
Dec 2 '18 at 23:19
$begingroup$
Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
$endgroup$
– lulu
Dec 2 '18 at 23:19
add a comment |
1 Answer
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$begingroup$
We first ignore the $A$'s and build the rest of the string. How many ways can we make a string of $6$ $B$'s and $3$ $C$'s such that the first $B$ comes before the first $C$? Well, the first letter has to be a $B$, and then anything goes after that; this gives $binom{8}{5}$ such strings ($8$ remaining letters to place, $5$ of which are $B$'s and $3$ of which are $C$'s).
Now we want to add $A$'s into this string. Note that the first letter has to be an $A$, and then the other $4$ $A$'s can be placed anywhere in our existing string. If we treat the $B$'s and $C$'s (which we've already fixed) as dividers, we see that this is really a stars-and-bars problem with $6+3+1$ categories (there are $9$ $B$'s and $C$'s, so $10$ "spots" to put $A$'s, including before/after all of them). So the number of ways to put $A$'s into the string is $binom{9 + 4}{4}$.
Putting these together, the total number of such strings is $binom{8}{5} binom{13}{4} = 40040$.
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We first ignore the $A$'s and build the rest of the string. How many ways can we make a string of $6$ $B$'s and $3$ $C$'s such that the first $B$ comes before the first $C$? Well, the first letter has to be a $B$, and then anything goes after that; this gives $binom{8}{5}$ such strings ($8$ remaining letters to place, $5$ of which are $B$'s and $3$ of which are $C$'s).
Now we want to add $A$'s into this string. Note that the first letter has to be an $A$, and then the other $4$ $A$'s can be placed anywhere in our existing string. If we treat the $B$'s and $C$'s (which we've already fixed) as dividers, we see that this is really a stars-and-bars problem with $6+3+1$ categories (there are $9$ $B$'s and $C$'s, so $10$ "spots" to put $A$'s, including before/after all of them). So the number of ways to put $A$'s into the string is $binom{9 + 4}{4}$.
Putting these together, the total number of such strings is $binom{8}{5} binom{13}{4} = 40040$.
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
We first ignore the $A$'s and build the rest of the string. How many ways can we make a string of $6$ $B$'s and $3$ $C$'s such that the first $B$ comes before the first $C$? Well, the first letter has to be a $B$, and then anything goes after that; this gives $binom{8}{5}$ such strings ($8$ remaining letters to place, $5$ of which are $B$'s and $3$ of which are $C$'s).
Now we want to add $A$'s into this string. Note that the first letter has to be an $A$, and then the other $4$ $A$'s can be placed anywhere in our existing string. If we treat the $B$'s and $C$'s (which we've already fixed) as dividers, we see that this is really a stars-and-bars problem with $6+3+1$ categories (there are $9$ $B$'s and $C$'s, so $10$ "spots" to put $A$'s, including before/after all of them). So the number of ways to put $A$'s into the string is $binom{9 + 4}{4}$.
Putting these together, the total number of such strings is $binom{8}{5} binom{13}{4} = 40040$.
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
We first ignore the $A$'s and build the rest of the string. How many ways can we make a string of $6$ $B$'s and $3$ $C$'s such that the first $B$ comes before the first $C$? Well, the first letter has to be a $B$, and then anything goes after that; this gives $binom{8}{5}$ such strings ($8$ remaining letters to place, $5$ of which are $B$'s and $3$ of which are $C$'s).
Now we want to add $A$'s into this string. Note that the first letter has to be an $A$, and then the other $4$ $A$'s can be placed anywhere in our existing string. If we treat the $B$'s and $C$'s (which we've already fixed) as dividers, we see that this is really a stars-and-bars problem with $6+3+1$ categories (there are $9$ $B$'s and $C$'s, so $10$ "spots" to put $A$'s, including before/after all of them). So the number of ways to put $A$'s into the string is $binom{9 + 4}{4}$.
Putting these together, the total number of such strings is $binom{8}{5} binom{13}{4} = 40040$.
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
$endgroup$
We first ignore the $A$'s and build the rest of the string. How many ways can we make a string of $6$ $B$'s and $3$ $C$'s such that the first $B$ comes before the first $C$? Well, the first letter has to be a $B$, and then anything goes after that; this gives $binom{8}{5}$ such strings ($8$ remaining letters to place, $5$ of which are $B$'s and $3$ of which are $C$'s).
Now we want to add $A$'s into this string. Note that the first letter has to be an $A$, and then the other $4$ $A$'s can be placed anywhere in our existing string. If we treat the $B$'s and $C$'s (which we've already fixed) as dividers, we see that this is really a stars-and-bars problem with $6+3+1$ categories (there are $9$ $B$'s and $C$'s, so $10$ "spots" to put $A$'s, including before/after all of them). So the number of ways to put $A$'s into the string is $binom{9 + 4}{4}$.
Putting these together, the total number of such strings is $binom{8}{5} binom{13}{4} = 40040$.
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
answered Dec 2 '18 at 23:30
plattyplatty
3,370320
3,370320
add a comment |
add a comment |
$begingroup$
Hint: a good arrangement must begin $A^nB$ for some $nin {1,2,3,4,5}$.
$endgroup$
– lulu
Dec 2 '18 at 23:19