anonymous
  • anonymous
Find the number of different arrangements that are possible for the letters. FACTOR
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Do i use permutation, combination, i don't get it
anonymous
  • anonymous
Are you kidding me?
anonymous
  • anonymous
How is this math?

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anonymous
  • anonymous
This is a brain teaser right?
anonymous
  • anonymous
No lol
anonymous
  • anonymous
its in the trigonometry curriculum
anonymous
  • anonymous
nope I have trigonometric nightmares during high school.
anonymous
  • anonymous
not for me. Biohazard
anonymous
  • anonymous
lol so you don't know how to do it?
anonymous
  • anonymous
Not a clue son
anonymous
  • anonymous
Im ultra retarded when it comes to anythign bu solving linear algebraic expressions or equations
anonymous
  • anonymous
Those I am good at
anonymous
  • 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?
anonymous
  • anonymous
Yes I am in college and the nightmare has just began...
anonymous
  • 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.
anonymous
  • 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 :)
anonymous
  • anonymous
@ujhk77 thats exactly what i got, just was unsure, thanks for the clarity, much appreciated :)
anonymous
  • anonymous
ok now that is a little better
anonymous
  • anonymous
@life No problemo bro, I'm happy to help. :)

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