Question

In how many ways can u arrange m black bead and n white bead in a necklace?

Answer 1

@experimentX PLZ try this if u r free.

Answer 2

probably we should work out on the other particular case ... generalizing these would be difficult.

Answer 3

i was working on that list .... could you correct it.
like in that 120 numbers, how many digits are repeated twice and how many are repeated thrice ... and how many are not repeating.

Answer 4

you want this in terms of m and n? its very similar to flipping heads in a row, with heads = black, tails = white etc

Answer 5

Yes interms of m and n

Answer 6

i will think about it :)

Answer 7

U mean it is like flipping a coin

Answer 8

yes sorry im tired, what list are you getting these problems from? sounds interesting.

Answer 9

http://openstudy.com/users/sauravshakya#/updates/507533f3e4b009782ca57f04

Answer 10

thanks

Answer 11

If we have a number x then in how many unique ways can we break it?

Answer 12

Suppose x=3
Then 3 , 1+2 , 1+1+1

Answer 13

I think if we can do that then we are not so far from the solution

Answer 14

okay ive been working on this for a while, im not sure how usefull this is but...
if s is the total amount of beads in the necklace, m is black beads, n is white beads and C is the amount of configurations then:
for n = s\[C = 2^s\]
for n = s -1\[C = 2^s - 1\]
for n = s - 2
\[C = 2^s -1 - s\]
for n = s - 3
\[C= 2^s - 1 - s - (s - 1) - (s - 2) - (s -3)\]

Answer 15

as you would expect this is symetrical so it works just as well for m as it does for n

Answer 16

also you can do it with n and m at the same time, so say you have a 4 beaded necklace, s = 4, and lets say you have a bag of 2 black and 3 white beads, m = 2 n = 3 so that n = s - 1 and m = s - 2. you will have.

\[C = 2^s - 1 - s - 1 \] configurations

Answer 17

obviously this is far from complete, as it only works fully for cases when n > s - 4 and m > s - 4.

Answer 18

Each time you add a new bead, you have 2 choices. So in total there will be \(2^{(m+k)}\) ways of arranging the beads.

Answer 19

Does that take into account the (circular) symmetry?

Answer 20

I guess yes

Answer 21

@Traxter I dont think that works

Answer 22

What I mean is that some patterns will be repeated due to the symmetry on the circle(neclace)

It's not the same as slaying them out in a line

Answer 23

Opps

Answer 24

Probably you would have to divide through by (m+n)....