Ask your own question, for FREE!
Mathematics 5 Online
OpenStudy (anonymous):

Find the number of different arrangements that are possible for the letters. FACTOR

OpenStudy (anonymous):

Do i use permutation, combination, i don't get it

OpenStudy (anonymous):

Are you kidding me?

OpenStudy (anonymous):

How is this math?

OpenStudy (anonymous):

This is a brain teaser right?

OpenStudy (anonymous):

No lol

OpenStudy (anonymous):

its in the trigonometry curriculum

OpenStudy (anonymous):

nope I have trigonometric nightmares during high school.

OpenStudy (anonymous):

not for me. Biohazard

OpenStudy (anonymous):

lol so you don't know how to do it?

OpenStudy (anonymous):

Not a clue son

OpenStudy (anonymous):

Im ultra retarded when it comes to anythign bu solving linear algebraic expressions or equations

OpenStudy (anonymous):

Those I am good at

OpenStudy (anonymous):

I was that person once lol, just gotta put some effort in, but from what you said earlier I'm guessing your done with high school, am i right?

OpenStudy (anonymous):

Yes I am in college and the nightmare has just began...

OpenStudy (anonymous):

@bunyonb It actually is math. There are six letters, none of which occur twice. The number of combinations for the letters can be found by a simple statistics formula: (x)! Where x is the number of elements that can be combined in a certain order. So, substitute 6 in for x to get: (6)! In case you're wondering the exclamation mark means factorial, which simply means that you multiply the integer by each of its preceding integers, all the way to one. So 6! = 6*5*4*3*2*1 = 720 Your answer is 720.

OpenStudy (anonymous):

lol, you seem like the same person as me, you don't wanna put in the effort, i could care less about math, but i still gotta do it, and your in college so its supper serious, good luck man :)

OpenStudy (anonymous):

@ujhk77 thats exactly what i got, just was unsure, thanks for the clarity, much appreciated :)

OpenStudy (anonymous):

ok now that is a little better

OpenStudy (anonymous):

@life No problemo bro, I'm happy to help. :)

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!