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Direct interpretation of $^9P_4times 2!$ as arrangement of $5$ people on $10$ seats in a row with two people together
Circular permutation - Arranging 4 persons around a circular table where 8 seats are there.arrangement of the word $“bfMATHEMATICS”$ in which no two same letter occur together.In how many ways if no two people of the same sex are allowed to sit together?Combinatorics - seating 7 people around table with 8 seats; two people have to be two seats apart7 seats in a row, probability of the married couple sitting togetherIf 4 people are seated randomly in a row of 8 seats, what is the probability that no 2 persons will sit on adjacent seats?In how many ways can you put two people in $5$ seats in a row if they cannot sit together?seating married couples around a circular tableProbability that in a row of 12 with random seating, two specific individuals are exactly two seats apartWays to arrange $ngeq2$ people around a circular table, given two permanent seats.
$begingroup$
To find the number of arrangements of $5$ people on $10$ seats in a row with $2$ particular people must be together, I know that the answer will be $$^9P_4times 2!$$
I have no problem on $2!$ because this is due to the arrangement of that two persons.
The problem I had is how to interpret the expression $^9P_4$ directly.
Previously I interpret it as:
I tie two persons that want to be together and "treat" them as one person. So combine with others this will give $4$ persons in total.
Next I tie two seats together and "treat" them as one seat. So combine with others this will give $9$ seats in total.
But I think that the way to interpret the seats is wrong since
(i) we do not know which two seats that we tie
(ii) we are arranging the persons on the seats not the other way round.
I know there are other ways to do this problem such as
(I) First of all arrange that two persons that want to be together, this will give $9$ possible ways. Then arrange the rest persons which give $^8P_3$, so we have $$9times ^8P_3= ^9P_4$$
(II) Arrange the persons together with the empty seats. Since empty seats are identical, this will give $$frac9!5!= ^9P_4$$
However, all these methods are not direct interpretation of $^9P_4$. Based on my understanding, $^9P_4$ is the arrangement of $4$ persons on $9$ seats, but I do not know how to interpret it in this problem.
combinatorics permutations
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
$endgroup$
add a comment |
$begingroup$
To find the number of arrangements of $5$ people on $10$ seats in a row with $2$ particular people must be together, I know that the answer will be $$^9P_4times 2!$$
I have no problem on $2!$ because this is due to the arrangement of that two persons.
The problem I had is how to interpret the expression $^9P_4$ directly.
Previously I interpret it as:
I tie two persons that want to be together and "treat" them as one person. So combine with others this will give $4$ persons in total.
Next I tie two seats together and "treat" them as one seat. So combine with others this will give $9$ seats in total.
But I think that the way to interpret the seats is wrong since
(i) we do not know which two seats that we tie
(ii) we are arranging the persons on the seats not the other way round.
I know there are other ways to do this problem such as
(I) First of all arrange that two persons that want to be together, this will give $9$ possible ways. Then arrange the rest persons which give $^8P_3$, so we have $$9times ^8P_3= ^9P_4$$
(II) Arrange the persons together with the empty seats. Since empty seats are identical, this will give $$frac9!5!= ^9P_4$$
However, all these methods are not direct interpretation of $^9P_4$. Based on my understanding, $^9P_4$ is the arrangement of $4$ persons on $9$ seats, but I do not know how to interpret it in this problem.
combinatorics permutations
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
$endgroup$
add a comment |
$begingroup$
To find the number of arrangements of $5$ people on $10$ seats in a row with $2$ particular people must be together, I know that the answer will be $$^9P_4times 2!$$
I have no problem on $2!$ because this is due to the arrangement of that two persons.
The problem I had is how to interpret the expression $^9P_4$ directly.
Previously I interpret it as:
I tie two persons that want to be together and "treat" them as one person. So combine with others this will give $4$ persons in total.
Next I tie two seats together and "treat" them as one seat. So combine with others this will give $9$ seats in total.
But I think that the way to interpret the seats is wrong since
(i) we do not know which two seats that we tie
(ii) we are arranging the persons on the seats not the other way round.
I know there are other ways to do this problem such as
(I) First of all arrange that two persons that want to be together, this will give $9$ possible ways. Then arrange the rest persons which give $^8P_3$, so we have $$9times ^8P_3= ^9P_4$$
(II) Arrange the persons together with the empty seats. Since empty seats are identical, this will give $$frac9!5!= ^9P_4$$
However, all these methods are not direct interpretation of $^9P_4$. Based on my understanding, $^9P_4$ is the arrangement of $4$ persons on $9$ seats, but I do not know how to interpret it in this problem.
combinatorics permutations
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
$endgroup$
To find the number of arrangements of $5$ people on $10$ seats in a row with $2$ particular people must be together, I know that the answer will be $$^9P_4times 2!$$
I have no problem on $2!$ because this is due to the arrangement of that two persons.
The problem I had is how to interpret the expression $^9P_4$ directly.
Previously I interpret it as:
I tie two persons that want to be together and "treat" them as one person. So combine with others this will give $4$ persons in total.
Next I tie two seats together and "treat" them as one seat. So combine with others this will give $9$ seats in total.
But I think that the way to interpret the seats is wrong since
(i) we do not know which two seats that we tie
(ii) we are arranging the persons on the seats not the other way round.
I know there are other ways to do this problem such as
(I) First of all arrange that two persons that want to be together, this will give $9$ possible ways. Then arrange the rest persons which give $^8P_3$, so we have $$9times ^8P_3= ^9P_4$$
(II) Arrange the persons together with the empty seats. Since empty seats are identical, this will give $$frac9!5!= ^9P_4$$
However, all these methods are not direct interpretation of $^9P_4$. Based on my understanding, $^9P_4$ is the arrangement of $4$ persons on $9$ seats, but I do not know how to interpret it in this problem.
combinatorics permutations
combinatorics permutations
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
edited Mar 22 at 9:28
Alan Wang
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
asked Mar 22 at 9:07
Alan WangAlan Wang
4,8921133
4,8921133
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Instead of "tying the seats", think about "tying the two people" that have to be together. Now you have 4 indivisible items that you have to put in 9 places ($^9P_4$ ways of doing). The "5th" individual and the "10th" chair will be placed either to the left of to the right next to their partner ($2!$ ways of doing).
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
$endgroup$
add a comment |
$begingroup$
Let's seat the pair of people first. There are nine ways to choose the leftmost seat occupied by the pair and two ways to arrange the people in the pair in those seats. That leaves eight seats for the remaining three people. They can be placed in those seats in $8 cdot 7 cdot 6$ ways. Hence, the number of admissible seating arrangements is $$9 cdot 2 cdot 8 cdot 7 cdot 6 = 9 cdot 8 cdot 7 cdot 6 cdot 2 = frac9 cdot 8 cdot 7 cdot 6 cdot 5!5! cdot 2! = frac9!5! cdot 2! = P(9, 4) cdot 2!$$
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
Instead of "tying the seats", think about "tying the two people" that have to be together. Now you have 4 indivisible items that you have to put in 9 places ($^9P_4$ ways of doing). The "5th" individual and the "10th" chair will be placed either to the left of to the right next to their partner ($2!$ ways of doing).
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
$endgroup$
add a comment |
$begingroup$
Instead of "tying the seats", think about "tying the two people" that have to be together. Now you have 4 indivisible items that you have to put in 9 places ($^9P_4$ ways of doing). The "5th" individual and the "10th" chair will be placed either to the left of to the right next to their partner ($2!$ ways of doing).
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
$endgroup$
add a comment |
$begingroup$
Instead of "tying the seats", think about "tying the two people" that have to be together. Now you have 4 indivisible items that you have to put in 9 places ($^9P_4$ ways of doing). The "5th" individual and the "10th" chair will be placed either to the left of to the right next to their partner ($2!$ ways of doing).
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
$endgroup$
Instead of "tying the seats", think about "tying the two people" that have to be together. Now you have 4 indivisible items that you have to put in 9 places ($^9P_4$ ways of doing). The "5th" individual and the "10th" chair will be placed either to the left of to the right next to their partner ($2!$ ways of doing).
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
answered Mar 22 at 9:43
ErtxiemErtxiem
667112
667112
add a comment |
add a comment |
$begingroup$
Let's seat the pair of people first. There are nine ways to choose the leftmost seat occupied by the pair and two ways to arrange the people in the pair in those seats. That leaves eight seats for the remaining three people. They can be placed in those seats in $8 cdot 7 cdot 6$ ways. Hence, the number of admissible seating arrangements is $$9 cdot 2 cdot 8 cdot 7 cdot 6 = 9 cdot 8 cdot 7 cdot 6 cdot 2 = frac9 cdot 8 cdot 7 cdot 6 cdot 5!5! cdot 2! = frac9!5! cdot 2! = P(9, 4) cdot 2!$$
$endgroup$
add a comment |
$begingroup$
Let's seat the pair of people first. There are nine ways to choose the leftmost seat occupied by the pair and two ways to arrange the people in the pair in those seats. That leaves eight seats for the remaining three people. They can be placed in those seats in $8 cdot 7 cdot 6$ ways. Hence, the number of admissible seating arrangements is $$9 cdot 2 cdot 8 cdot 7 cdot 6 = 9 cdot 8 cdot 7 cdot 6 cdot 2 = frac9 cdot 8 cdot 7 cdot 6 cdot 5!5! cdot 2! = frac9!5! cdot 2! = P(9, 4) cdot 2!$$
$endgroup$
add a comment |
$begingroup$
Let's seat the pair of people first. There are nine ways to choose the leftmost seat occupied by the pair and two ways to arrange the people in the pair in those seats. That leaves eight seats for the remaining three people. They can be placed in those seats in $8 cdot 7 cdot 6$ ways. Hence, the number of admissible seating arrangements is $$9 cdot 2 cdot 8 cdot 7 cdot 6 = 9 cdot 8 cdot 7 cdot 6 cdot 2 = frac9 cdot 8 cdot 7 cdot 6 cdot 5!5! cdot 2! = frac9!5! cdot 2! = P(9, 4) cdot 2!$$
$endgroup$
Let's seat the pair of people first. There are nine ways to choose the leftmost seat occupied by the pair and two ways to arrange the people in the pair in those seats. That leaves eight seats for the remaining three people. They can be placed in those seats in $8 cdot 7 cdot 6$ ways. Hence, the number of admissible seating arrangements is $$9 cdot 2 cdot 8 cdot 7 cdot 6 = 9 cdot 8 cdot 7 cdot 6 cdot 2 = frac9 cdot 8 cdot 7 cdot 6 cdot 5!5! cdot 2! = frac9!5! cdot 2! = P(9, 4) cdot 2!$$
answered Mar 22 at 9:49
community wiki
N. F. Taussig
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