1. ## How many different

There are 20 questions in the multichoice-exam and there are 4 answer alternatives on every question.
Only one alternative is always correct.
How many different answer papers is possible to form in which there are 70% of the answers correct?
Here is my solution ( I am not sure is this same formula needed here, but it might be right ) and that 70% would be * 0,7

(20 + 4 - 1)!
__________ * 0,7 =
20! * ( 4 - 1)!

2. ## Re: How many different

Originally Posted by stevetall
There are 20 questions in the multichoice-exam and there are 4 answer alternatives on every question.
Only one alternative is always correct.
How many different answer papers is possible to form in which there are 70% of the answers correct?
There are $\displaystyle \binom{20}{14}$ ways to have exactly fourteen correct answers.
There are $\displaystyle 3^6$ ways to answer the other six incorrectly.

Thank you.

4. ## Re: How many different

I just want to make sure that I understood.

So this is 20 over 14

20
14

and should there be 4 *, I mean;

20 * 4
14

Because each question contains 4 alternatives. This probably is a big number.
It cannot be * 4! , but it might be * 4

5. ## Re: How many different

Originally Posted by stevetall
So this is 20 over 14
$\displaystyle \binom{20}{14}=\frac{20!}{14!\cdot 6!}=38760$

6. ## Re: How many different

Ah, of course...

Now it is clear.

Thanks!