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Life

  • 2 years ago

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

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  1. Life
    • 2 years ago
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    Do i use permutation, combination, i don't get it

  2. bunyonb
    • 2 years ago
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    Are you kidding me?

  3. bunyonb
    • 2 years ago
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    How is this math?

  4. bunyonb
    • 2 years ago
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    This is a brain teaser right?

  5. Life
    • 2 years ago
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    No lol

  6. Life
    • 2 years ago
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    its in the trigonometry curriculum

  7. bunyonb
    • 2 years ago
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    nope I have trigonometric nightmares during high school.

  8. bunyonb
    • 2 years ago
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    not for me. Biohazard

  9. Life
    • 2 years ago
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    lol so you don't know how to do it?

  10. bunyonb
    • 2 years ago
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    Not a clue son

  11. bunyonb
    • 2 years ago
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    Im ultra retarded when it comes to anythign bu solving linear algebraic expressions or equations

  12. bunyonb
    • 2 years ago
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    Those I am good at

  13. Life
    • 2 years ago
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    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?

  14. bunyonb
    • 2 years ago
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    Yes I am in college and the nightmare has just began...

  15. ujhk77
    • 2 years ago
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    @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.

  16. Life
    • 2 years ago
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    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 :)

  17. Life
    • 2 years ago
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    @ujhk77 thats exactly what i got, just was unsure, thanks for the clarity, much appreciated :)

  18. bunyonb
    • 2 years ago
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    ok now that is a little better

  19. ujhk77
    • 2 years ago
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    @life No problemo bro, I'm happy to help. :)

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