Combinatorics - How many orders are possible, from a chef's and waiter's perspective? [closed]
$begingroup$
A family of $7$ are out to lunch and each of them orders one dish from a menu of $12$ dishes.
- (a) From the waiter’s perspective, how many orders are possible?
- (b) From the chef’s perspective, how many orders are possible?
- (c) What is the effect of changing the perspective in the above?
I tried $_{12}C_7$ for both (a) & (b), but I don't understand what's the difference between each perspective. Aren't they the same?
combinatorics permutations combinations
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
$endgroup$
closed as off-topic by Namaste, A. Pongrácz, Holo, mrtaurho, user91500 Dec 31 '18 at 11:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Holo, mrtaurho, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A family of $7$ are out to lunch and each of them orders one dish from a menu of $12$ dishes.
- (a) From the waiter’s perspective, how many orders are possible?
- (b) From the chef’s perspective, how many orders are possible?
- (c) What is the effect of changing the perspective in the above?
I tried $_{12}C_7$ for both (a) & (b), but I don't understand what's the difference between each perspective. Aren't they the same?
combinatorics permutations combinations
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
$endgroup$
closed as off-topic by Namaste, A. Pongrácz, Holo, mrtaurho, user91500 Dec 31 '18 at 11:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Holo, mrtaurho, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34
add a comment |
$begingroup$
A family of $7$ are out to lunch and each of them orders one dish from a menu of $12$ dishes.
- (a) From the waiter’s perspective, how many orders are possible?
- (b) From the chef’s perspective, how many orders are possible?
- (c) What is the effect of changing the perspective in the above?
I tried $_{12}C_7$ for both (a) & (b), but I don't understand what's the difference between each perspective. Aren't they the same?
combinatorics permutations combinations
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
$endgroup$
A family of $7$ are out to lunch and each of them orders one dish from a menu of $12$ dishes.
- (a) From the waiter’s perspective, how many orders are possible?
- (b) From the chef’s perspective, how many orders are possible?
- (c) What is the effect of changing the perspective in the above?
I tried $_{12}C_7$ for both (a) & (b), but I don't understand what's the difference between each perspective. Aren't they the same?
combinatorics permutations combinations
combinatorics permutations combinations
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
edited Dec 28 '18 at 3:42
Eevee Trainer
8,16521439
8,16521439
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
asked Dec 28 '18 at 3:22
llamaro25llamaro25
528
528
closed as off-topic by Namaste, A. Pongrácz, Holo, mrtaurho, user91500 Dec 31 '18 at 11:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Holo, mrtaurho, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, A. Pongrácz, Holo, mrtaurho, user91500 Dec 31 '18 at 11:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Holo, mrtaurho, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34
add a comment |
2
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34
2
2
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the chef, it’s the number of solutions to $n_1+n_2+ dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18choose7$. For the waiter, it’s $12^7$.
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
add a comment |
$begingroup$
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$begin{align}
text{# of permutations with reptition} &= n^k\
text{# of combinations with reptition} &= _{n+k-1}C_k = binom{n+k-1}{k} = frac{(n+k-1)!}{k! (n-1)!}\
end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
$endgroup$
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the chef, it’s the number of solutions to $n_1+n_2+ dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18choose7$. For the waiter, it’s $12^7$.
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
add a comment |
$begingroup$
For the chef, it’s the number of solutions to $n_1+n_2+ dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18choose7$. For the waiter, it’s $12^7$.
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
$endgroup$
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
add a comment |
$begingroup$
For the chef, it’s the number of solutions to $n_1+n_2+ dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18choose7$. For the waiter, it’s $12^7$.
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
$endgroup$
For the chef, it’s the number of solutions to $n_1+n_2+ dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18choose7$. For the waiter, it’s $12^7$.
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
answered Dec 28 '18 at 3:41
Steve KassSteve Kass
11.4k11530
11.4k11530
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
add a comment |
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
Thanks for the solution. I get it now :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
$begingroup$
It’s called “stars and bars.” To count the ways, count the number of ways to write* * | * | | | * * | | |... with *s for the meals, first group is item 1 orders, etc., with 11 separators meaning “switch to next menu item.”
$endgroup$
– Steve Kass
Dec 28 '18 at 3:48
add a comment |
$begingroup$
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$begin{align}
text{# of permutations with reptition} &= n^k\
text{# of combinations with reptition} &= _{n+k-1}C_k = binom{n+k-1}{k} = frac{(n+k-1)!}{k! (n-1)!}\
end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
$endgroup$
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
add a comment |
$begingroup$
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$begin{align}
text{# of permutations with reptition} &= n^k\
text{# of combinations with reptition} &= _{n+k-1}C_k = binom{n+k-1}{k} = frac{(n+k-1)!}{k! (n-1)!}\
end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
$endgroup$
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
add a comment |
$begingroup$
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$begin{align}
text{# of permutations with reptition} &= n^k\
text{# of combinations with reptition} &= _{n+k-1}C_k = binom{n+k-1}{k} = frac{(n+k-1)!}{k! (n-1)!}\
end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
$endgroup$
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$begin{align}
text{# of permutations with reptition} &= n^k\
text{# of combinations with reptition} &= _{n+k-1}C_k = binom{n+k-1}{k} = frac{(n+k-1)!}{k! (n-1)!}\
end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
edited Dec 28 '18 at 3:44
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
answered Dec 28 '18 at 3:29
Eevee TrainerEevee Trainer
8,16521439
8,16521439
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
add a comment |
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
$begingroup$
Oh, I get it. Thanks :)
$endgroup$
– llamaro25
Dec 28 '18 at 3:47
add a comment |
2
$begingroup$
The waiter will care who ordered which dish. The chef will not.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:24
$begingroup$
@farruhota Slight problem with that (I fell into the same hole and deleted my answer): what if two people order the same meal? Those formulas don't account for such repetition.
$endgroup$
– Eevee Trainer
Dec 28 '18 at 3:34