Question

Combination question

Answer 1

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?

Answer 2

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!

Answer 3

Why do we divide by @! @Vocaloid? I don't get that :/

Answer 4

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

Answer 5

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

Answer 6

[small typo in the first post, meant to say 3! instead of 6!]

Answer 7

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?

Answer 8

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!)

Answer 9

so, in the denominator, you need to include all the repeated letters

Answer 10

WHOOPS, forgot that :P
So do we multiple the repetitions, like 3! times 2!, or add them?

Answer 11

multiply

Answer 12

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

Answer 13

well, there's always the formula

nPr = n!/(n-r)!

Answer 14

I mean, how do we do that if we could have repetitions?

Answer 15

can you clarify what you mean?

Answer 16

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

Answer 17

Does that make sense @Vocaloid?

Answer 18

I understand the question, I'm just a bit uncertain on the answer

Answer 19

intuitively I would think

8P3/(3!2!)

I'll get a second opinion though @ganeshie8

Answer 20

I thought that aswell. Thanks so much @Vocaloid :D

Answer 21

@Vocaloid, did you get it in the end?

Answer 22

no response yet, sorry D:

I need to go to sleep, tag someone if you still want a second opinion

Answer 23

That's cool. Thanks anyway @Vocaloid :D