Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

The high school dance team is holding auditions to fill the spots of four members that graduated. If 10 people try out for the 4 positions, and order does not matter, how many possible combinations exist?

hero (hero):

I created a little chart for these at one point, but lost it

OpenStudy (anonymous):

aww

hero (hero):

Let me see if I can find it again

hero (hero):

Here's the website I used to create the charts. I must have been a genius back then because I don't even remember it anymore

hero (hero):

It's like you consider four things: Combination Permutation Order Repetition

OpenStudy (anonymous):

Order doesn't matter..... could it be 10C4 ?\[\left(\begin{matrix}10 \\ 4\end{matrix}\right)=\frac{10!}{4!6!}=\frac{10*9*8*7*6!}{4*3*2*6!}=210\]

hero (hero):

And based on that, there is a formula

OpenStudy (anonymous):

i was just about to do that formula :)

hero (hero):

We know it is combinations and we know order doesn't matter, so there is a formula that corresponds with that

hero (hero):

But there's like several formulas for different scenarios

hero (hero):

I'd have to put them all together again.

hero (hero):

Did you find all four of them?

hero (hero):

If not, I can post them again

OpenStudy (anonymous):

Animalani actually got the answer i just had to find 1

OpenStudy (anonymous):

and it corresponded with the 4 answers that were given thanks for your help though

hero (hero):

@AnimalAin just like posting answers.

hero (hero):

Giving answers never really helps anyone understand anything though

OpenStudy (anonymous):

I thought I showed the justification and work for it, as well....

hero (hero):

You did, but you explained nothing.

OpenStudy (anonymous):

Hero i found the formula out while animalain was typing so either way we would have came to the same conclusion

OpenStudy (anonymous):

so it somewhat helped me i guess

hero (hero):

@jazzy2346 , I'm trying to give @AnimalAin a hard time either way.

OpenStudy (anonymous):

lol

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!