$10$ students are randomly seated in a row. $3$ of these students are brothers. Find

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$10$ students are randomly seated in a row. $3$ of these students are brothers.



a) What is the probability that these students will sit next to each other?




The way I pictured it , b b b s s s s s s s 



 1 2 3 4 5 6 7 8


grouped the brothers as 1.



Therefore, $dfrac8!3!10! = dfrac115$.



The $3!$ is the ways to rearrange the $3$ boys.




b) What is the probability that exactly two of them will sit next to each other?




b b s b s s s s s s



s b b s s s b s s s



All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.



$$fracbinom32210!$$




probability




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    Please read this tutorial on how to typeset mathematics on this site.
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    Jul 20 at 7:37














up vote
1
down vote

favorite













$10$ students are randomly seated in a row. $3$ of these students are brothers.



a) What is the probability that these students will sit next to each other?




The way I pictured it , b b b s s s s s s s 



 1 2 3 4 5 6 7 8


grouped the brothers as 1.



Therefore, $dfrac8!3!10! = dfrac115$.



The $3!$ is the ways to rearrange the $3$ boys.




b) What is the probability that exactly two of them will sit next to each other?




b b s b s s s s s s



s b b s s s b s s s



All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.



$$fracbinom32210!$$




probability




share|cite|improve this question

















  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Jul 20 at 7:37












up vote
1
down vote

favorite









up vote
1
down vote

favorite












$10$ students are randomly seated in a row. $3$ of these students are brothers.



a) What is the probability that these students will sit next to each other?




The way I pictured it , b b b s s s s s s s 



 1 2 3 4 5 6 7 8


grouped the brothers as 1.



Therefore, $dfrac8!3!10! = dfrac115$.



The $3!$ is the ways to rearrange the $3$ boys.




b) What is the probability that exactly two of them will sit next to each other?




b b s b s s s s s s



s b b s s s b s s s



All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.



$$fracbinom32210!$$




probability




share|cite|improve this question














$10$ students are randomly seated in a row. $3$ of these students are brothers.



a) What is the probability that these students will sit next to each other?




The way I pictured it , b b b s s s s s s s 



 1 2 3 4 5 6 7 8


grouped the brothers as 1.



Therefore, $dfrac8!3!10! = dfrac115$.



The $3!$ is the ways to rearrange the $3$ boys.




b) What is the probability that exactly two of them will sit next to each other?




b b s b s s s s s s



s b b s s s b s s s



All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.



$$fracbinom32210!$$





probability





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 20 at 7:36









N. F. Taussig

38.2k93053




38.2k93053









asked Jul 20 at 4:12









Sacha E

376




376







  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Jul 20 at 7:37












  • 1




    Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    Jul 20 at 7:37







1




1




Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Jul 20 at 7:37




Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
Jul 20 at 7:37










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










(a) Yes. There are $8!cdot 3!$ ways to arrange the ten individuals when the brothers are grouped three-together; your argument for this is sound.   Also a total of $10!$ ways to arrange the individuals with no such constraint.



Alternatively you could count ways to: arrange the three brothers, arrange the seven others, then select one from the eight spaces between-or-outside them to place the line of brothers. $$3!~7!~binom 81$$



Which, of course, has the same value.



(b) Use that alternative method.   Count ways to arrange the seven others, then ways to select, arrange, and place the brothers so that they are a pair and a singleton seperated by at least one other.




All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.




$checkmark$.   You have counted ways to selected and arranged the pair.   Now count the ways to arrange the seven others, place the pair in one from eight spaces, and place the lone brother in one from the remainder.






share|cite|improve this answer



















  • 1




    Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
    – Sacha E
    Jul 20 at 17:10






  • 1




    $checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
    – Graham Kemp
    Jul 21 at 2:01










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










(a) Yes. There are $8!cdot 3!$ ways to arrange the ten individuals when the brothers are grouped three-together; your argument for this is sound.   Also a total of $10!$ ways to arrange the individuals with no such constraint.



Alternatively you could count ways to: arrange the three brothers, arrange the seven others, then select one from the eight spaces between-or-outside them to place the line of brothers. $$3!~7!~binom 81$$



Which, of course, has the same value.



(b) Use that alternative method.   Count ways to arrange the seven others, then ways to select, arrange, and place the brothers so that they are a pair and a singleton seperated by at least one other.




All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.




$checkmark$.   You have counted ways to selected and arranged the pair.   Now count the ways to arrange the seven others, place the pair in one from eight spaces, and place the lone brother in one from the remainder.






share|cite|improve this answer



















  • 1




    Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
    – Sacha E
    Jul 20 at 17:10






  • 1




    $checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
    – Graham Kemp
    Jul 21 at 2:01














up vote
2
down vote



accepted










(a) Yes. There are $8!cdot 3!$ ways to arrange the ten individuals when the brothers are grouped three-together; your argument for this is sound.   Also a total of $10!$ ways to arrange the individuals with no such constraint.



Alternatively you could count ways to: arrange the three brothers, arrange the seven others, then select one from the eight spaces between-or-outside them to place the line of brothers. $$3!~7!~binom 81$$



Which, of course, has the same value.



(b) Use that alternative method.   Count ways to arrange the seven others, then ways to select, arrange, and place the brothers so that they are a pair and a singleton seperated by at least one other.




All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.




$checkmark$.   You have counted ways to selected and arranged the pair.   Now count the ways to arrange the seven others, place the pair in one from eight spaces, and place the lone brother in one from the remainder.






share|cite|improve this answer



















  • 1




    Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
    – Sacha E
    Jul 20 at 17:10






  • 1




    $checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
    – Graham Kemp
    Jul 21 at 2:01












up vote
2
down vote



accepted







up vote
2
down vote



accepted






(a) Yes. There are $8!cdot 3!$ ways to arrange the ten individuals when the brothers are grouped three-together; your argument for this is sound.   Also a total of $10!$ ways to arrange the individuals with no such constraint.



Alternatively you could count ways to: arrange the three brothers, arrange the seven others, then select one from the eight spaces between-or-outside them to place the line of brothers. $$3!~7!~binom 81$$



Which, of course, has the same value.



(b) Use that alternative method.   Count ways to arrange the seven others, then ways to select, arrange, and place the brothers so that they are a pair and a singleton seperated by at least one other.




All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.




$checkmark$.   You have counted ways to selected and arranged the pair.   Now count the ways to arrange the seven others, place the pair in one from eight spaces, and place the lone brother in one from the remainder.






share|cite|improve this answer















(a) Yes. There are $8!cdot 3!$ ways to arrange the ten individuals when the brothers are grouped three-together; your argument for this is sound.   Also a total of $10!$ ways to arrange the individuals with no such constraint.



Alternatively you could count ways to: arrange the three brothers, arrange the seven others, then select one from the eight spaces between-or-outside them to place the line of brothers. $$3!~7!~binom 81$$



Which, of course, has the same value.



(b) Use that alternative method.   Count ways to arrange the seven others, then ways to select, arrange, and place the brothers so that they are a pair and a singleton seperated by at least one other.




All I got so far is I must pick $2$ of those $3$, and $2!$ ways to rearrange them.




$checkmark$.   You have counted ways to selected and arranged the pair.   Now count the ways to arrange the seven others, place the pair in one from eight spaces, and place the lone brother in one from the remainder.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 20 at 8:19


























answered Jul 20 at 5:20









Graham Kemp

80.1k43275




80.1k43275







  • 1




    Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
    – Sacha E
    Jul 20 at 17:10






  • 1




    $checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
    – Graham Kemp
    Jul 21 at 2:01












  • 1




    Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
    – Sacha E
    Jul 20 at 17:10






  • 1




    $checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
    – Graham Kemp
    Jul 21 at 2:01







1




1




Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
– Sacha E
Jul 20 at 17:10




Ohhhhhhh, so would it be [(3)(2)(7!)(8)(7)]/(10!)?. select 2 out of 3 brothers, ways to arrange. 7! ways to arrange the non brothers, 8c1 ways to place the two brothers with the non brothers, and 7 ways to add the last brother so he is not next to the two others.
– Sacha E
Jul 20 at 17:10




1




1




$checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
– Graham Kemp
Jul 21 at 2:01




$checkmark$ Otherwise written as: $^3mathrm P_2^7mathrm P_7^8mathrm P_2/^10mathrm P_10$
– Graham Kemp
Jul 21 at 2:01












 

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