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Find the number of ways in which a list can be formed of the order of the 24 boats
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There are 15 rowing clubs;two of the clubs have each 3 boats on the river;5 others have each 2 and the remaining eight have each 1;find the number of ways in which a list can be formed of the order of the 24 boats,observing that the second boat of a club cannot be above the first and the third above the second.How many ways are there in which a boat of the club having single boat on the river is at the third place in the list formed above?
The number of ways in which a list can be formed of the order of the 24 boats $=24!$ but i cannot interpret what the second boat of a club cannot be above the first and the third above the second means.
The answers given is $frac24!(3!)^2(2!)^5,binom81frac23!(3!)^2(2!)^5.$
combinatorics
asked Jul 18 at 8:15
learner_avid
682413
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There are 15 rowing clubs;two of the clubs have each 3 boats on the river;5 others have each 2 and the remaining eight have each 1;find the number of ways in which a list can be formed of the order of the 24 boats,observing that the second boat of a club cannot be above the first and the third above the second.How many ways are there in which a boat of the club having single boat on the river is at the third place in the list formed above?
The number of ways in which a list can be formed of the order of the 24 boats $=24!$ but i cannot interpret what the second boat of a club cannot be above the first and the third above the second means.
The answers given is $frac24!(3!)^2(2!)^5,binom81frac23!(3!)^2(2!)^5.$
combinatorics
asked Jul 18 at 8:15
learner_avid
682413
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There are 15 rowing clubs;two of the clubs have each 3 boats on the river;5 others have each 2 and the remaining eight have each 1;find the number of ways in which a list can be formed of the order of the 24 boats,observing that the second boat of a club cannot be above the first and the third above the second.How many ways are there in which a boat of the club having single boat on the river is at the third place in the list formed above?
The number of ways in which a list can be formed of the order of the 24 boats $=24!$ but i cannot interpret what the second boat of a club cannot be above the first and the third above the second means.
The answers given is $frac24!(3!)^2(2!)^5,binom81frac23!(3!)^2(2!)^5.$
combinatorics
asked Jul 18 at 8:15
learner_avid
682413
There are 15 rowing clubs;two of the clubs have each 3 boats on the river;5 others have each 2 and the remaining eight have each 1;find the number of ways in which a list can be formed of the order of the 24 boats,observing that the second boat of a club cannot be above the first and the third above the second.How many ways are there in which a boat of the club having single boat on the river is at the third place in the list formed above?
The number of ways in which a list can be formed of the order of the 24 boats $=24!$ but i cannot interpret what the second boat of a club cannot be above the first and the third above the second means.
The answers given is $frac24!(3!)^2(2!)^5,binom81frac23!(3!)^2(2!)^5.$
combinatorics
asked Jul 18 at 8:15
learner_avid
682413
asked Jul 18 at 8:15
learner_avid
682413
asked Jul 18 at 8:15
learner_avid
682413
asked Jul 18 at 8:15
learner_avid
682413
682413
add a comment |Â
add a comment |Â
2 Answers
2
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2
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At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8timesfrac23!3!^22!^5$$ orderings that satisfy all conditions.
answered Jul 18 at 8:41
drhab
86.5k541118
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HINT: You can interpret the question as the following:
In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
answered Jul 18 at 8:31
ArsenBerk
6,3021932
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8timesfrac23!3!^22!^5$$ orderings that satisfy all conditions.
answered Jul 18 at 8:41
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
accepted
At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8timesfrac23!3!^22!^5$$ orderings that satisfy all conditions.
answered Jul 18 at 8:41
drhab
86.5k541118
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8timesfrac23!3!^22!^5$$ orderings that satisfy all conditions.
answered Jul 18 at 8:41
drhab
86.5k541118
At first a boat is chosen from one of the $8$ clubs that only have a single boat, and this boat gets number $3$ on the list. This can be done in $8$ ways.
Then $23$ boats are left and if they are placed unconditionally that gives $23!$ possibilities. Looking at the conditions for a club that has $3$ boats we find only $1$ on the $3!=6$ orderings satisfies the condition. Now looking at the conditions for a club that has $2$ boats we find only $1$ on the $2!=2$ orderings satisfies the condition.
Repairing this gives a total of: $$8timesfrac23!3!^22!^5$$ orderings that satisfy all conditions.
answered Jul 18 at 8:41
drhab
86.5k541118
answered Jul 18 at 8:41
drhab
86.5k541118
answered Jul 18 at 8:41
drhab
86.5k541118
answered Jul 18 at 8:41
drhab
86.5k541118
86.5k541118
add a comment |Â
add a comment |Â
up vote
1
down vote
HINT: You can interpret the question as the following:
In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
answered Jul 18 at 8:31
ArsenBerk
6,3021932
add a comment |Â
up vote
1
down vote
HINT: You can interpret the question as the following:
In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
answered Jul 18 at 8:31
ArsenBerk
6,3021932
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT: You can interpret the question as the following:
In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
answered Jul 18 at 8:31
ArsenBerk
6,3021932
HINT: You can interpret the question as the following:
In the image above, suppose we have a stack of piles and we can't take the pile $3$ without taking $1$ and $2$; and we can't take the pile $2$ without taking $1$ (because pile $3$ is under pile $1$ and $2$ and pile $2$ is under pile $1$). In this case, notice that there is only $1$ way of putting them to another rod (in order of $3-2-1$ from top to bottom).
In your question, the logic is the same. Suppose we have stacks of numbered piles (from $1$ to $24$) and piles are distributed so that there are two stacks with $3$ piles, five stacks with $2$ piles and eight stacks with $1$ pile. Then we are trying to put these $24$ piles into a single rod.
answered Jul 18 at 8:31
ArsenBerk
6,3021932
answered Jul 18 at 8:31
ArsenBerk
6,3021932
answered Jul 18 at 8:31
ArsenBerk
6,3021932
answered Jul 18 at 8:31
ArsenBerk
6,3021932
6,3021932
add a comment |Â
add a comment |Â
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