Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

On a 20 problem test, how many ways can a person get 4 problems correct and 16 problems wrong?

Sagot :

This is a combination problem where the order doesn't matter.

We compute the number of combinations with the following formula:

[tex]C^{n_{}}_r=\frac{n!}{r!(n-r)!}[/tex]

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

In our problem, we have n = 20 and for r we can pick r = 16 or r =4 (with both options we get the same result), picking r = 16 we get:

[tex]C^{20_{}}_{16}=\frac{20!}{16!\cdot(20-16)!}=\frac{20!}{16!\cdot4!}=4845[/tex]

Answer

We have 4845 ways possible to get 4 problems correct and 16 problems wrong from a 20 problem set.

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.