Ahsome
  • Ahsome
Combination question
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Ahsome
  • Ahsome
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?
Vocaloid
  • Vocaloid
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!
Ahsome
  • Ahsome
Why do we divide by @! @Vocaloid? I don't get that :/

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Vocaloid
  • Vocaloid
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
Vocaloid
  • Vocaloid
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
Vocaloid
  • Vocaloid
[small typo in the first post, meant to say 3! instead of 6!]
Ahsome
  • Ahsome
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?
Vocaloid
  • Vocaloid
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!)
Vocaloid
  • Vocaloid
so, in the denominator, you need to include all the repeated letters
Ahsome
  • Ahsome
WHOOPS, forgot that :P So do we multiple the repetitions, like 3! times 2!, or add them?
Vocaloid
  • Vocaloid
multiply
Ahsome
  • Ahsome
Thank you so much! :D How would this work for Permutations tho, where R and N aren't the same number? @Vocaloid?
Vocaloid
  • Vocaloid
well, there's always the formula nPr = n!/(n-r)!
Ahsome
  • Ahsome
I mean, how do we do that if we could have repetitions?
Vocaloid
  • Vocaloid
can you clarify what you mean?
Ahsome
  • Ahsome
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
Ahsome
  • Ahsome
Does that make sense @Vocaloid?
Vocaloid
  • Vocaloid
I understand the question, I'm just a bit uncertain on the answer
Vocaloid
  • Vocaloid
intuitively I would think 8P3/(3!2!) I'll get a second opinion though @ganeshie8
Ahsome
  • Ahsome
I thought that aswell. Thanks so much @Vocaloid :D
Ahsome
  • Ahsome
@Vocaloid, did you get it in the end?
Vocaloid
  • Vocaloid
no response yet, sorry D: I need to go to sleep, tag someone if you still want a second opinion
Ahsome
  • Ahsome
That's cool. Thanks anyway @Vocaloid :D

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