Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Can somebody help me re-learn combinations and permutations? I haven't used them in years and I'm taking the GRE in the next month!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

do you have a specific question? it is easy enough to write a formula, but that might not be helpful

- anonymous

Well, I don't have any specific questions. It's just that I can't just memorize the formulas without fully understanding them.

- anonymous

ok well if you want "permutations" like arranging 4 people in 4 chairs, the number of ways to do it is
\[4!=4\times 3\times 2\] and that is simple enough to understand by the "counting principle"
4 choices for the first seat
3 for the second
2 for the third
and no choice for the last one

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Yeah, I understand that.. But then once I get to combinations, I get confused.

- anonymous

http://www.khanacademy.org/video/permutations?playlist=Precalculus

- anonymous

Seriously. Go there.

- anonymous

Watch that video and the ones that follow.

- anonymous

oh ok well if you understand that, then we are on our way. suppose then you want to know how many committees of 4 people you can make out of a total of ten people. then again by the counting principle you have
\[10\times 9\times 8\times 7\] ways to do it, but that counts too many because it counts all the permutations as different. so you have to divide by the number of ways you can permute the 4 people, and we just said what that way, namely 4!

- anonymous

so the answer would be
\[\frac{10\times 9\times 8\times 7}{4\times 3\times 2}\]

- anonymous

ok

- anonymous

what about if repetitions are allowed?

Looking for something else?

Not the answer you are looking for? Search for more explanations.