In how many ways can a necklace with $18$ different beads be cut into two necklaces with at least six beads on each?

Question

I have a circular necklace with $18$ beads on it. All the beads are different. Making two cuts with a pair of scissors, I can divide the necklace into two strings of beads. If I want each string to have at least $6$ beads, how many different pairs of strings can I make?

Answer 1

(You were right, I corrected my answer.)

There are four general types you can make: $(6,12), (7,11), (8,10), (9,9).$

First type: $(6,12)$, there are $18$ possible ways to choose those $6$, which are going to be cut from the necklace: choose a direction to count the beads, and choose a starting position (between $2$ beads). There are exactly $18$ starting positions, since there are $18$ spaces between the beads. One cut is equivalent for a pair you can make.

Second type: $(7,11)$, with the same reasoning, there are $18$ possible ways to cut the necklace.

Third type: $(8,10)$, $18$ again.

Fourth type: $(9,9)$, The same reasoning cannot be applied again, since half of the cuts would be exactly the same as the other half. So there are $9$ possible cuts, exactly one or each axis of symmetry for the necklace.

Edit: The solution will be the sum these values: $18+18+18+9=63.$

Answer 2

The circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is $\frac12(n-1)!$

In this problem, you need to have at least $6$ beads.

So,

$$=\frac12(6-1)!+\frac12(7-1)!+\frac12(8-1)!+\frac12(9-1)!$$

Since you can divide the necklace into two strings of beads,

$$=2(\frac12(6-1)!+\frac12(7-1)!+\frac12(8-1)!+\frac12(9-1)!)$$

Answer 3

There are $18$ possible positions for the first cut. After we make the first cut, we have a string of $18$ beads in a row. There are $7$ places we can make the second cut (it can be after the $6^{\rm th}$, $7^{\rm th}$, $8^{\rm th}$, $9^{\rm th}$, $10^{\rm th}$, $11^{\rm th}$, or $12^{\rm th}$ bead). So, we have $18\cdot 7$ choices for the two cuts. However, it doesn't matter which cut we made first and which cut we made second, so we've counted every possible end result twice. Thus, the number of pairs of strings we can make is $(18\cdot 7)/2$, which is $\boxed{63}$.