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Ahsome

  • one year ago

Combination question

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  1. ahsome
    • one year ago
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    In how many ways can the letters of the word NEWTON be arranged if they are used once only and taken 6 at a time, assuming, there is no distinction between the two Ns?

  2. Vocaloid
    • one year ago
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    well, newton has 6 letters so we start with 6! to account for the 2 duplicate n's, we divide by 2! so our answer is 6!/2!

  3. ahsome
    • one year ago
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    Why do we divide by @! @Vocaloid? I don't get that :/

  4. Vocaloid
    • one year ago
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    well, that's a bit tough to explain, maybe I can use a simpler example, let's say ABC and AAC for ABC, our possible arrangements are ABC, ACB, BAC, BCA, CAB, CBA, giving us 6 arrangements, or 6! for AAC, I can use the same pattern and get 6 arrangements, but because we have two A's, some of those arrangements are duplicated. our arrangements are AAC, ACA, AAC, ACA, CAA, CAA. there are 6 total, but only 3 of them are unique arrangements. we can get this result mathematically by dividing 3!/2! which gives us 6/2 = 3

  5. Vocaloid
    • one year ago
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    so, basically, what I'm getting at is: if we have duplicate letters, you will end up getting repeat arrangements that get counted multiple times. in order to account for these, we divide by (number of repeat letters)! for each repeated letter I'm not great at explaining things, please let me know if you are still confused @Ahsome

  6. Vocaloid
    • one year ago
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    [small typo in the first post, meant to say 3! instead of 6!]

  7. ahsome
    • one year ago
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    I NOTICED :p No no no, that makes sense :) So, like if we had the question that we had the word PARALLEL. Do we use: \[\dfrac{8!}{3!}\]Cause the letter L is repeated 3 times?

  8. Vocaloid
    • one year ago
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    almost! you have a repeated L 3 times, but you also have a repeated letter A two times, so we do: 8! divided by (3!2!)

  9. Vocaloid
    • one year ago
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    so, in the denominator, you need to include all the repeated letters

  10. ahsome
    • one year ago
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    WHOOPS, forgot that :P So do we multiple the repetitions, like 3! times 2!, or add them?

  11. Vocaloid
    • one year ago
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    multiply

  12. ahsome
    • one year ago
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    Thank you so much! :D How would this work for Permutations tho, where R and N aren't the same number? @Vocaloid?

  13. Vocaloid
    • one year ago
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    well, there's always the formula nPr = n!/(n-r)!

  14. ahsome
    • one year ago
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    I mean, how do we do that if we could have repetitions?

  15. Vocaloid
    • one year ago
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    can you clarify what you mean?

  16. ahsome
    • one year ago
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    Like, imagine I had the word PARALLAL What are the combinations if I used those words to make a 3 letter word, without any repeats

  17. ahsome
    • one year ago
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    Does that make sense @Vocaloid?

  18. Vocaloid
    • one year ago
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    I understand the question, I'm just a bit uncertain on the answer

  19. Vocaloid
    • one year ago
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    intuitively I would think 8P3/(3!2!) I'll get a second opinion though @ganeshie8

  20. ahsome
    • one year ago
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    I thought that aswell. Thanks so much @Vocaloid :D

  21. ahsome
    • one year ago
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    @Vocaloid, did you get it in the end?

  22. Vocaloid
    • one year ago
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    no response yet, sorry D: I need to go to sleep, tag someone if you still want a second opinion

  23. ahsome
    • one year ago
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    That's cool. Thanks anyway @Vocaloid :D

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