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Statistics in permutations and combinations [closed]

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0

$begingroup$

A student must answer 8 out of 10 questions in an exam. (1) In how many ways can this be done if he/she is free to choose any 8 of the 10? (2) In how many ways can this be done if he/she must answer at least 4 of the first 5 questions?

combinatorics permutations

share|cite|improve this question

edited Mar 10 at 22:31

Sjava Mabuza

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

$endgroup$

closed as off-topic by Eevee Trainer, darij grinberg, pwerth, Parcly Taxel, Leucippus Mar 11 at 4:01

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." – Eevee Trainer, darij grinberg, pwerth, Parcly Taxel, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  Mar 10 at 22:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (2) separate into two mutually exclusive cases: - answering exactly 4 of the first 5 questions and answering exactly 5 of the first 5 questions
  $endgroup$
  – WW1
  Mar 10 at 22:32
  

  
  
  
  

add a comment | 

0

$begingroup$

A student must answer 8 out of 10 questions in an exam. (1) In how many ways can this be done if he/she is free to choose any 8 of the 10? (2) In how many ways can this be done if he/she must answer at least 4 of the first 5 questions?

combinatorics permutations

share|cite|improve this question

edited Mar 10 at 22:31

Sjava Mabuza

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

$endgroup$

closed as off-topic by Eevee Trainer, darij grinberg, pwerth, Parcly Taxel, Leucippus Mar 11 at 4:01

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." – Eevee Trainer, darij grinberg, pwerth, Parcly Taxel, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  Mar 10 at 22:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (2) separate into two mutually exclusive cases: - answering exactly 4 of the first 5 questions and answering exactly 5 of the first 5 questions
  $endgroup$
  – WW1
  Mar 10 at 22:32
  

  
  
  
  

add a comment | 

$begingroup$

A student must answer 8 out of 10 questions in an exam. (1) In how many ways can this be done if he/she is free to choose any 8 of the 10? (2) In how many ways can this be done if he/she must answer at least 4 of the first 5 questions?

combinatorics permutations

share|cite|improve this question

edited Mar 10 at 22:31

Sjava Mabuza

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

$endgroup$

share|cite|improve this question

edited Mar 10 at 22:31

Sjava Mabuza

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

share|cite|improve this question

edited Mar 10 at 22:31

Sjava Mabuza

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

asked Mar 10 at 22:19

Sjava MabuzaSjava Mabuza

82

1 Answer
1

active

oldest

votes

0

$begingroup$

Let $A$ represent an answered question, then $U$ represent an unanswered question.
For the first $5$ questions, you have $5$ places you can "put" the $U$.
$$UAAAA$$ $$AUAAA$$ $$AAUAA$$ $$AAAUA$$ $$AAAAU$$
For the remaining $5$ questions you have to answer $4$, so the pattern is the same. So you have $5$ ways in the first half, and $5$ ways in the second half... this is a compound probability.

share|cite|improve this answer

answered Mar 10 at 22:31

CarserCarser

2,63541027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well... you changed the question after I started answering it.
  $endgroup$
  – Carser
  Mar 10 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So what exactly is the solution?
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know how probability works?
  $endgroup$
  – Carser
  Mar 10 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I know how it works
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! The above is the solution.
  $endgroup$
  – Carser
  Mar 10 at 22:36
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Let $A$ represent an answered question, then $U$ represent an unanswered question.
For the first $5$ questions, you have $5$ places you can "put" the $U$.
$$UAAAA$$ $$AUAAA$$ $$AAUAA$$ $$AAAUA$$ $$AAAAU$$
For the remaining $5$ questions you have to answer $4$, so the pattern is the same. So you have $5$ ways in the first half, and $5$ ways in the second half... this is a compound probability.

share|cite|improve this answer

answered Mar 10 at 22:31

CarserCarser

2,63541027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well... you changed the question after I started answering it.
  $endgroup$
  – Carser
  Mar 10 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So what exactly is the solution?
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know how probability works?
  $endgroup$
  – Carser
  Mar 10 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I know how it works
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! The above is the solution.
  $endgroup$
  – Carser
  Mar 10 at 22:36
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

Let $A$ represent an answered question, then $U$ represent an unanswered question.
For the first $5$ questions, you have $5$ places you can "put" the $U$.
$$UAAAA$$ $$AUAAA$$ $$AAUAA$$ $$AAAUA$$ $$AAAAU$$
For the remaining $5$ questions you have to answer $4$, so the pattern is the same. So you have $5$ ways in the first half, and $5$ ways in the second half... this is a compound probability.

share|cite|improve this answer

answered Mar 10 at 22:31

CarserCarser

2,63541027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well... you changed the question after I started answering it.
  $endgroup$
  – Carser
  Mar 10 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So what exactly is the solution?
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know how probability works?
  $endgroup$
  – Carser
  Mar 10 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I know how it works
  $endgroup$
  – Sjava Mabuza
  Mar 10 at 22:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! The above is the solution.
  $endgroup$
  – Carser
  Mar 10 at 22:36
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Let $A$ represent an answered question, then $U$ represent an unanswered question.
For the first $5$ questions, you have $5$ places you can "put" the $U$.
$$UAAAA$$ $$AUAAA$$ $$AAUAA$$ $$AAAUA$$ $$AAAAU$$
For the remaining $5$ questions you have to answer $4$, so the pattern is the same. So you have $5$ ways in the first half, and $5$ ways in the second half... this is a compound probability.

share|cite|improve this answer

answered Mar 10 at 22:31

CarserCarser

2,63541027

$endgroup$

share|cite|improve this answer

answered Mar 10 at 22:31

CarserCarser

2,63541027

answered Mar 10 at 22:31

CarserCarser

2,63541027

answered Mar 10 at 22:31

CarserCarser

2,63541027