General stats question (true/false test)

[Linty Fresh]

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]

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]

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]

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]

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]

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]

Of course! Thanks. I'm getting the formulas down, but still kind of a beginner in the practical application.