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How many ways could the guests arrange themselves on a four person couch
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There are 33 guest at your home for a dinner party in how many ways could the guests arrange themselves on a four person couch
I really not understand this question form where to start can someone explain please
thanks ..
combinatorics permutations combinations
asked 1 hour ago
learner
515
add a comment |Â
up vote
3
down vote
favorite
There are 33 guest at your home for a dinner party in how many ways could the guests arrange themselves on a four person couch
I really not understand this question form where to start can someone explain please
thanks ..
combinatorics permutations combinations
asked 1 hour ago
learner
515
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There are 33 guest at your home for a dinner party in how many ways could the guests arrange themselves on a four person couch
I really not understand this question form where to start can someone explain please
thanks ..
combinatorics permutations combinations
asked 1 hour ago
learner
515
There are 33 guest at your home for a dinner party in how many ways could the guests arrange themselves on a four person couch
I really not understand this question form where to start can someone explain please
thanks ..
combinatorics permutations combinations
combinatorics permutations combinations
asked 1 hour ago
learner
515
asked 1 hour ago
learner
515
asked 1 hour ago
learner
515
asked 1 hour ago
learner
515
asked 1 hour ago
learner
515
515
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
From the 33 guests, choose 4 to sit on the four-person couch: $33 choose 4$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is:
$$
33 choose 4 4!
$$
answered 1 hour ago
Bruno Reis
750416
add a comment |Â
up vote
2
down vote
This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n geq r$.
You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^st$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^nd$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^rd$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^th$ box, where there will be $(n-(r-1))$ options.
Let’s say $P$ represents the total number of permutations for $n$ choose $r$.
$$P_(n, r) = ncdot(n-1)cdot(n-2)cdot(n-3)...cdot(n-(r-1)) = fracn!(n-r)!$$
In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$.
$$P_(33, 4) = frac33!(33-4)! = frac33!29!$$
All factors from $29$ to $1$ will cancel.
$$P_(33, 4) = 33cdot 32cdot 31cdot 30$$
$$boxedP_(33, 4) = 982080$$
answered 49 mins ago
KM101
894110
add a comment |Â
up vote
1
down vote
If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.
On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...
So $33*32*31*30=982080$ ways.
answered 1 hour ago
MathFun123
1489
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
From the 33 guests, choose 4 to sit on the four-person couch: $33 choose 4$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is:
$$
33 choose 4 4!
$$
answered 1 hour ago
Bruno Reis
750416
add a comment |Â
up vote
3
down vote
accepted
From the 33 guests, choose 4 to sit on the four-person couch: $33 choose 4$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is:
$$
33 choose 4 4!
$$
answered 1 hour ago
Bruno Reis
750416
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
From the 33 guests, choose 4 to sit on the four-person couch: $33 choose 4$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is:
$$
33 choose 4 4!
$$
answered 1 hour ago
Bruno Reis
750416
From the 33 guests, choose 4 to sit on the four-person couch: $33 choose 4$. After that, arrange those 4 guests on the couch in all possible ways: $4!$. Hence the final answer is:
$$
33 choose 4 4!
$$
answered 1 hour ago
Bruno Reis
750416
answered 1 hour ago
Bruno Reis
750416
answered 1 hour ago
Bruno Reis
750416
answered 1 hour ago
Bruno Reis
750416
750416
add a comment |Â
add a comment |Â
up vote
2
down vote
This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n geq r$.
You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^st$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^nd$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^rd$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^th$ box, where there will be $(n-(r-1))$ options.
Let’s say $P$ represents the total number of permutations for $n$ choose $r$.
$$P_(n, r) = ncdot(n-1)cdot(n-2)cdot(n-3)...cdot(n-(r-1)) = fracn!(n-r)!$$
In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$.
$$P_(33, 4) = frac33!(33-4)! = frac33!29!$$
All factors from $29$ to $1$ will cancel.
$$P_(33, 4) = 33cdot 32cdot 31cdot 30$$
$$boxedP_(33, 4) = 982080$$
answered 49 mins ago
KM101
894110
add a comment |Â
up vote
2
down vote
This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n geq r$.
You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^st$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^nd$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^rd$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^th$ box, where there will be $(n-(r-1))$ options.
Let’s say $P$ represents the total number of permutations for $n$ choose $r$.
$$P_(n, r) = ncdot(n-1)cdot(n-2)cdot(n-3)...cdot(n-(r-1)) = fracn!(n-r)!$$
In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$.
$$P_(33, 4) = frac33!(33-4)! = frac33!29!$$
All factors from $29$ to $1$ will cancel.
$$P_(33, 4) = 33cdot 32cdot 31cdot 30$$
$$boxedP_(33, 4) = 982080$$
answered 49 mins ago
KM101
894110
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n geq r$.
You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^st$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^nd$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^rd$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^th$ box, where there will be $(n-(r-1))$ options.
Let’s say $P$ represents the total number of permutations for $n$ choose $r$.
$$P_(n, r) = ncdot(n-1)cdot(n-2)cdot(n-3)...cdot(n-(r-1)) = fracn!(n-r)!$$
In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$.
$$P_(33, 4) = frac33!(33-4)! = frac33!29!$$
All factors from $29$ to $1$ will cancel.
$$P_(33, 4) = 33cdot 32cdot 31cdot 30$$
$$boxedP_(33, 4) = 982080$$
answered 49 mins ago
KM101
894110
This is a case of permutations without repetition. To visualize this better, imagine you have a total of $n$ balls and you want to place them in $r$ boxes. Each box can contain only $1$ ball. Clearly, the number of boxes filled cannot surpass $n$, since $n geq r$.
You want to see the number of different ways you could do that. In other words, you want to find the number of permutations. For the $1^st$ box, you could place any one of the balls. Hence, there are $n$ options. For the $2^nd$ box, you have $1$ fewer ball, so there will be $(n-1)$ options. For the $3^rd$ box, there will be $(n-2)$ options. Keep repeating until you reach the $r^th$ box, where there will be $(n-(r-1))$ options.
Let’s say $P$ represents the total number of permutations for $n$ choose $r$.
$$P_(n, r) = ncdot(n-1)cdot(n-2)cdot(n-3)...cdot(n-(r-1)) = fracn!(n-r)!$$
In your question, there are $33$ guests and a $4$-person couch, so $n = 33$ and $r = 4$.
$$P_(33, 4) = frac33!(33-4)! = frac33!29!$$
All factors from $29$ to $1$ will cancel.
$$P_(33, 4) = 33cdot 32cdot 31cdot 30$$
$$boxedP_(33, 4) = 982080$$
answered 49 mins ago
KM101
894110
answered 49 mins ago
KM101
894110
answered 49 mins ago
KM101
894110
answered 49 mins ago
KM101
894110
894110
add a comment |Â
add a comment |Â
up vote
1
down vote
If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.
On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...
So $33*32*31*30=982080$ ways.
answered 1 hour ago
MathFun123
1489
add a comment |Â
up vote
1
down vote
If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.
On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...
So $33*32*31*30=982080$ ways.
answered 1 hour ago
MathFun123
1489
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.
On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...
So $33*32*31*30=982080$ ways.
answered 1 hour ago
MathFun123
1489
If I understand the question correctly; we have one couch for four people and 33 guests so 29 guests have to stand.
On the first place on the couch 33 people can sit, on the second place only 32 can sit (because one is already sitting on the first place), and so on...
So $33*32*31*30=982080$ ways.
answered 1 hour ago
MathFun123
1489
answered 1 hour ago
MathFun123
1489
answered 1 hour ago
MathFun123
1489
answered 1 hour ago
MathFun123
1489
1489
add a comment |Â
add a comment |Â
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