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General stats question (true/false test)

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
I'm trying to teach myself stats and probability, and I've been doing practice problems. A problem just occurred to me, and I thought I'd run it by the board.

Take a twenty question True/False test. How many ways can you answer ten questions true and ten false?

How would I go about starting to solve this? Would I start with the nPr formula or try to visualize the different ways you can arrange the trues and falses?

Many thanks!
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,708
1) Start Small and get your hands dirty.

How about 2 with 1 and 1?
How about 4 with 2 and 2?

Does one lead to the other? Is there a relationship?

How about 6 with 3 and 3?

Did anything you learn from the first two surprise you that it didn't work, here? Is there something new? Can you chop this down so that it's jsut a step harder than 4 with 2 and 2?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, Linty Fresh!

Take a twenty-question True/False test.
How many ways can you answer ten questions true and ten false?

This is a "Combinations" problem.

Select 10 of the 20 questions and answer them "True".
(Of course, the other ten questions will be answered "False".)

\(\displaystyle \text{The number of ways is: }\:_{20}C_{10} \:=\:{20\choose10} \:=\:\frac{20!}{10!\,10!} \:=\: 184,\!756\)

 

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
hi soroban,
I'm sorry to bother you but could you check my message?
I did ask you something. please let me know if you could.
many thanks~
 

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
soroban said:
Hello, Linty Fresh!

Take a twenty-question True/False test.
How many ways can you answer ten questions true and ten false?

This is a "Combinations" problem.

Select 10 of the 20 questions and answer them "True".
(Of course, the other ten questions will be answered "False".)

\(\displaystyle \text{The number of ways is: }\:_{20}C_{10} \:=\:{20\choose10} \:=\:\frac{20!}{10!\,10!} \:=\: 184,\!756\)

OK, really dumb question: Why wouldn't it be a permutations problem, that is 20!/10!. Why divide r out at all?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, Linty Fresh!

OK, really dumb question: Why wouldn't it be a permutations problem?

\(\displaystyle \text{A permutation imparts an }order\text{ to the selections.}\)

\(\displaystyle \text{Suppose you choose to answer True to questions: }\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)

\(\displaystyle \text{Using permutations, }\{2,1,3,4,5,6,7,8,9,10\}\text{ is a different choice.}\)

 

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
Of course! Thanks. I'm getting the formulas down, but still kind of a beginner in the practical application.
 
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