On a(n) 25-problem test, how many ways can a person get 8 problems correct and 17 problems wrong?

Question

On a(n) 25-problem test, how many ways can a person get 8 problems correct and 17 problems wrong?

Answer 1

When the order doesn't matter, it is a Combination problem. (1,2,3 is the same as 3,2,1 or 2,3,1)

It's also important to decide whether repeat solutions (3,3,3) are allowed which they're not in this case as a student cannot get the same question correct more than once.

 

nCr(n,r) = n!/(r!(n-r)!

 

Where n is the number of things to choose from, and we choose r of them. (No repetition, order doesn't matter)

 

nCr(25,8) = 25!/(8!(25-8)!)

 

nCr(25,8) = 1,081,575