Question

Find the number of necklaces that can be made Usin 3 beads of one kind and 9 beads of the other kind.

Answer 1

Is it possible to generalize the above obtained result?

Answer 2

(9! 3!)/2

Answer 3

That is not correct.

Answer 4

@experimentX

Answer 5

circular permutation ,,, have to check for circular symmetry.

Answer 6

Why do you say so?

Answer 7

As in the 18.

Answer 8

lol ... can't be 18 ... should be pretty close though

Answer 9

It is.
But what is the best way to solve for this?
And, how'd you approximate it, man?

Answer 10

lol .. really? ... all i know is it should be pretty close to this figure.
\[ \binom{12}{3} \over 12\]

Answer 11

Okay. Yeah. Makes sense.
Then,?

Answer 12

I've seen this problem before ... i know a bit rigorous method. counting
111000000000 ---- 1 <-- for this type

101100000000 ---- 10-3 = 7 <-- check out for symmetry (incorrect)

Answer 13

100001100000 <--- only 4 due to circular symmetry.

Answer 14

101010000000 <-- for this type ... only 1

Answer 15

looks like we should be hunting down asymmetric cases.

Answer 16

for all symmetric cases we will have 1 ... let's check out symmetric cases.

Answer 17

What is this youre doing again. In the sense what is 101010 mean?

Answer 18

1 <-- type 1
0 <-- type 2

Answer 19

Okay.

Answer 20

the figure can be more than 18 (might be ...)

Answer 21

symmetric cases
111000000000
101010000000
100100100000
100010001000
seems only three symmetric cases.

Answer 22

Isnt brute force not a very good way.
I dunno. It might work for this but we could never generalize it which is the next part of the question.

Answer 23

honestly ... i don't know. let's try using brute force for now ... something might happen
asymmetric cases
101100000000
101010000000
101000100000
101000010000 <--- from the other side ... this is enough

Answer 24

101100000000 < 4
101010000000 < 3
101000100000 < 2
101000010000 < 1

where are the other cases missing?

Answer 25

101100000000 < 5
101010000000 < 4
101001000000 < 3
101000100000 < 2
101000010000 < 1

Answer 26

with 3 symmetric cases we have total of 18 permutations. looks like we can generalize.

Answer 27

@siddhantsharan you still there?

Answer 28

let's try this will 3 type 1 beads and 10 type 2 beads.

Answer 29

Okay. I'm back. Sorry.

Answer 30

Oh ... i'm back too.

Answer 31

Hmm.
How about this.
Let the 3 beads be like: