Why did we multiply (b) by 5! and 6!?

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Six people enter an elevator at the ground floor of a 12-storey building. They choose their exit floors independently of each other.

Why did we multiply (b) by 5! and 6!?
Does the order matters? If so, why didn't we take into consideration the order in (a).

probability

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asked Jul 23 at 14:13

Rayri

485

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In both parts, it does matter who gets off the elevator on which floor.
  – N. F. Taussig
  Jul 23 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So why then, we did not multiply by 5! And 6! In a)?
  – Rayri
  Jul 23 at 14:25
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Six people enter an elevator at the ground floor of a 12-storey building. They choose their exit floors independently of each other.

Why did we multiply (b) by 5! and 6!?
Does the order matters? If so, why didn't we take into consideration the order in (a).

probability

share|cite|improve this question

asked Jul 23 at 14:13

Rayri

485

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In both parts, it does matter who gets off the elevator on which floor.
  – N. F. Taussig
  Jul 23 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So why then, we did not multiply by 5! And 6! In a)?
  – Rayri
  Jul 23 at 14:25
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Six people enter an elevator at the ground floor of a 12-storey building. They choose their exit floors independently of each other.

Why did we multiply (b) by 5! and 6!?
Does the order matters? If so, why didn't we take into consideration the order in (a).

probability

share|cite|improve this question

asked Jul 23 at 14:13

Rayri

485

Six people enter an elevator at the ground floor of a 12-storey building. They choose their exit floors independently of each other.

Why did we multiply (b) by 5! and 6!?
Does the order matters? If so, why didn't we take into consideration the order in (a).

probability

share|cite|improve this question

asked Jul 23 at 14:13

Rayri

485

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 14:13

Rayri

485

asked Jul 23 at 14:13

Rayri

485

asked Jul 23 at 14:13

Rayri

485

485

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In both parts, it does matter who gets off the elevator on which floor.
  – N. F. Taussig
  Jul 23 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So why then, we did not multiply by 5! And 6! In a)?
  – Rayri
  Jul 23 at 14:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In both parts, it does matter who gets off the elevator on which floor.
  – N. F. Taussig
  Jul 23 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So why then, we did not multiply by 5! And 6! In a)?
  – Rayri
  Jul 23 at 14:25
  

  
  
  
  

In both parts, it does matter who gets off the elevator on which floor.
– N. F. Taussig
Jul 23 at 14:22

In both parts, it does matter who gets off the elevator on which floor.
– N. F. Taussig
Jul 23 at 14:22

So why then, we did not multiply by 5! And 6! In a)?
– Rayri
Jul 23 at 14:25

So why then, we did not multiply by 5! And 6! In a)?
– Rayri
Jul 23 at 14:25

add a comment | 

3 Answers
3

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oldest

votes

up vote
1
down vote



accepted

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $binom122$ ways of selecting the two floors at which two people each exit the elevator and $binom102$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $binom62$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $binom42$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $binom21$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is
$$binom122binom102binom62binom42binom21$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is
$$frac111 cdot frac12 = frac122$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $binom116$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is
$$binom1166!left(frac122right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $binom115$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is
$$binom61left(frac12right)binom115left(frac122right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

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answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

add a comment | 

up vote
2
down vote

The solution may have been more transparent if instead of writing what they did, they wrote

$$
_11P_6left(frac122right)^6 + left(6 times frac12right) left[ _11P_5 left(frac122right)^5right]
$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
  
  
  • $_11P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
  
  
  • $1/22$ is the probability of getting out at any non-top floor
  
  


• Case II: One person gets off on the top floor
  
  
  • $6$ is the number of ways to select the person who gets out on the top floor
  
  
  • $1/2$ is the probability of getting out on the top floor
  
  
  • $_11P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
  
  
  • $1/22$ is again the probability of getting out on any non-top floor
  
  

Note that

$$
_nP_k = k!_nC_k = k!binomnk
$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

---

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

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edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

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up vote
2
down vote

In part a) they're still taking into account who gets off at which floor. There are $12 choose 2$ ways to pick the two floors at which two people get off and $10choose2$ ways to choose the floors at which single people get off. Then there are $6choose2$ ways to assign two people to the first "double" floor, $4choose2$ ways to assign two people to the second double floor, and $2choose1$ ways to assign two people to the "single" floors.

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answered Jul 23 at 14:56

saulspatz

10.5k21323

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $binom122$ ways of selecting the two floors at which two people each exit the elevator and $binom102$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $binom62$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $binom42$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $binom21$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is
$$binom122binom102binom62binom42binom21$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is
$$frac111 cdot frac12 = frac122$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $binom116$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is
$$binom1166!left(frac122right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $binom115$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is
$$binom61left(frac12right)binom115left(frac122right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

share|cite|improve this answer

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

add a comment | 

up vote
1
down vote



accepted

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $binom122$ ways of selecting the two floors at which two people each exit the elevator and $binom102$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $binom62$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $binom42$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $binom21$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is
$$binom122binom102binom62binom42binom21$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is
$$frac111 cdot frac12 = frac122$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $binom116$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is
$$binom1166!left(frac122right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $binom115$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is
$$binom61left(frac12right)binom115left(frac122right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

share|cite|improve this answer

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $binom122$ ways of selecting the two floors at which two people each exit the elevator and $binom102$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $binom62$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $binom42$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $binom21$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is
$$binom122binom102binom62binom42binom21$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is
$$frac111 cdot frac12 = frac122$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $binom116$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is
$$binom1166!left(frac122right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $binom115$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is
$$binom61left(frac12right)binom115left(frac122right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

share|cite|improve this answer

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Find the probability that exactly two people get out on one floor, exactly two people on another floor, and each of the remaining people gets off where no one else does.

Since each of the six people can choose to exit the elevator on one of the twelve floors, there are $12^6$ elements in the sample space.

There are $binom122$ ways of selecting the two floors at which two people each exit the elevator and $binom102$ ways of selecting at which two of the remaining ten floors one person each exits the elevator. There are $binom62$ ways of selecting which two people exit the elevator on the lower of the two floors from which two people exit the elevator and $binom42$ ways of selecting which two of the remaining four people exit the elevator on the higher of those floors. There are $binom21$ ways of selecting which one of the remaining two people exits the elevator on the lower of the two floors at which one person exits the elevator. The remaining person must exit the elevator on the higher of the two floors at which one person exits the elevator. Hence, the number of favorable cases is
$$binom122binom102binom62binom42binom21$$

Notice that it matters which person exits the elevator on which floor.

Six people enter an elevator on the ground floor of a $12$-storey building. They choose their exit floors independently of each other. Assume instead each person exits the top floor with probability $1/2$ and with equal probability at each of the other floors. Determine the probability that they all get out at different floors.

It does matter who exits the elevator on which floor.

Since the probability that a person exits the elevator on the top floor is $1/2$, the probability that she or he does not exit the elevator on the top floor is $1 - 1/2 = 1/2$. Since it is equally likely that the person exits the elevator on any of those eleven floors, the probability that a person exits at a particular lower floor is
$$frac111 cdot frac12 = frac122$$

There are two possibilities for the favorable cases. Either nobody gets off the elevator on the top floor or one person does.

We find the probabilities of each case.

Case 1: Nobody exits at the top floor.

There are $binom116$ ways to select at which six of the eleven lower floors the six people exit the elevator and $6!$ ways to assign them to those floors. Each of those six people has probability of $1/22$ of exiting the elevator at that particular floor. Since these choices are independent, the probability that six people exit the elevator at six different floors, none of which is the top floor, is
$$binom1166!left(frac122right)^6$$

Case 2: One person exits at the top floor.

There are six ways to select the person who exits the elevator at the top floor. The probability that he or she exits the elevator on that floor is $1/2$. Each of the other people must exit the elevator at a different one of the lower eleven floors. There are $binom115$ ways to select the five floors at which they exit and $5!$ ways to arrange them on those floors. Each of those five people has probability $1/22$ of selecting that particular floor. Since these choices are independent, the probability that each person exits the elevator at a different floor if one of them chooses to exit at the top floor is
$$binom61left(frac12right)binom115left(frac122right)^5$$

Since these cases are mutually exclusive, the desired probability is obtained by adding the results for the two cases.

share|cite|improve this answer

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

answered Jul 23 at 14:59

N. F. Taussig

38.2k93053

38.2k93053

add a comment | 

add a comment | 

up vote
2
down vote

The solution may have been more transparent if instead of writing what they did, they wrote

$$
_11P_6left(frac122right)^6 + left(6 times frac12right) left[ _11P_5 left(frac122right)^5right]
$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
  
  
  • $_11P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
  
  
  • $1/22$ is the probability of getting out at any non-top floor
  
  


• Case II: One person gets off on the top floor
  
  
  • $6$ is the number of ways to select the person who gets out on the top floor
  
  
  • $1/2$ is the probability of getting out on the top floor
  
  
  • $_11P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
  
  
  • $1/22$ is again the probability of getting out on any non-top floor
  
  

Note that

$$
_nP_k = k!_nC_k = k!binomnk
$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

---

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

share|cite|improve this answer

edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

add a comment | 

up vote
2
down vote

The solution may have been more transparent if instead of writing what they did, they wrote

$$
_11P_6left(frac122right)^6 + left(6 times frac12right) left[ _11P_5 left(frac122right)^5right]
$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
  
  
  • $_11P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
  
  
  • $1/22$ is the probability of getting out at any non-top floor
  
  


• Case II: One person gets off on the top floor
  
  
  • $6$ is the number of ways to select the person who gets out on the top floor
  
  
  • $1/2$ is the probability of getting out on the top floor
  
  
  • $_11P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
  
  
  • $1/22$ is again the probability of getting out on any non-top floor
  
  

Note that

$$
_nP_k = k!_nC_k = k!binomnk
$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

---

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

share|cite|improve this answer

edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

add a comment | 

up vote
2
down vote

up vote
2
down vote

The solution may have been more transparent if instead of writing what they did, they wrote

$$
_11P_6left(frac122right)^6 + left(6 times frac12right) left[ _11P_5 left(frac122right)^5right]
$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
  
  
  • $_11P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
  
  
  • $1/22$ is the probability of getting out at any non-top floor
  
  


• Case II: One person gets off on the top floor
  
  
  • $6$ is the number of ways to select the person who gets out on the top floor
  
  
  • $1/2$ is the probability of getting out on the top floor
  
  
  • $_11P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
  
  
  • $1/22$ is again the probability of getting out on any non-top floor
  
  

Note that

$$
_nP_k = k!_nC_k = k!binomnk
$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

---

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

share|cite|improve this answer

edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

The solution may have been more transparent if instead of writing what they did, they wrote

$$
_11P_6left(frac122right)^6 + left(6 times frac12right) left[ _11P_5 left(frac122right)^5right]
$$

for that is equal to the expression in (b). From left to write, we have

• Case I: No one gets off on the top floor
  
  
  • $_11P_6$ is the number of permutations of $6$ out of the $11$ non-top floors
  
  
  • $1/22$ is the probability of getting out at any non-top floor
  
  


• Case II: One person gets off on the top floor
  
  
  • $6$ is the number of ways to select the person who gets out on the top floor
  
  
  • $1/2$ is the probability of getting out on the top floor
  
  
  • $_11P_5$ is the number of permutations of $5$ out of the $11$ non-top floors
  
  
  • $1/22$ is again the probability of getting out on any non-top floor
  
  

Note that

$$
_nP_k = k!_nC_k = k!binomnk
$$

We take permutations—in other words, the order matters here—because assigning (for instance) floor $5$ to person $1$ and floor $7$ to person $2$, etc., is different from assigning floor $7$ to person $1$ and floor $5$ to person $2$, etc.

---

I'm still trying to parse the answer to (a). I have to say, I'm afraid, their presentation is not the best. ETA: OK, I've now got it, but I see that saulspatz has anticipated me and explained part (a) nicely.

share|cite|improve this answer

edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 14:56

edited Jul 23 at 14:56

edited Jul 23 at 14:56

answered Jul 23 at 14:51

Brian Tung

25.2k32453

answered Jul 23 at 14:51

Brian Tung

25.2k32453

answered Jul 23 at 14:51

Brian Tung

25.2k32453

25.2k32453

add a comment | 

add a comment | 

up vote
2
down vote

In part a) they're still taking into account who gets off at which floor. There are $12 choose 2$ ways to pick the two floors at which two people get off and $10choose2$ ways to choose the floors at which single people get off. Then there are $6choose2$ ways to assign two people to the first "double" floor, $4choose2$ ways to assign two people to the second double floor, and $2choose1$ ways to assign two people to the "single" floors.

share|cite|improve this answer

answered Jul 23 at 14:56

saulspatz

10.5k21323

add a comment | 

up vote
2
down vote

In part a) they're still taking into account who gets off at which floor. There are $12 choose 2$ ways to pick the two floors at which two people get off and $10choose2$ ways to choose the floors at which single people get off. Then there are $6choose2$ ways to assign two people to the first "double" floor, $4choose2$ ways to assign two people to the second double floor, and $2choose1$ ways to assign two people to the "single" floors.

share|cite|improve this answer

answered Jul 23 at 14:56

saulspatz

10.5k21323

add a comment | 

up vote
2
down vote

up vote
2
down vote

In part a) they're still taking into account who gets off at which floor. There are $12 choose 2$ ways to pick the two floors at which two people get off and $10choose2$ ways to choose the floors at which single people get off. Then there are $6choose2$ ways to assign two people to the first "double" floor, $4choose2$ ways to assign two people to the second double floor, and $2choose1$ ways to assign two people to the "single" floors.

share|cite|improve this answer

answered Jul 23 at 14:56

saulspatz

10.5k21323

In part a) they're still taking into account who gets off at which floor. There are $12 choose 2$ ways to pick the two floors at which two people get off and $10choose2$ ways to choose the floors at which single people get off. Then there are $6choose2$ ways to assign two people to the first "double" floor, $4choose2$ ways to assign two people to the second double floor, and $2choose1$ ways to assign two people to the "single" floors.

share|cite|improve this answer

answered Jul 23 at 14:56

saulspatz

10.5k21323

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 14:56

saulspatz

10.5k21323

answered Jul 23 at 14:56

saulspatz

10.5k21323

answered Jul 23 at 14:56

saulspatz

10.5k21323

10.5k21323

add a comment | 

add a comment | 

 

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