Kitty has 5 small, 3 medium and 2 large spherical charms and each of the 10 charms is uniquely designed.

1. Treat each size group as a single block and arrange the 3 blocks in a circle.

There are 3 groups of charms (small, medium, large).
The number of ways to arrange 3 groups in a circle is $(3 - 1)! = 2! = 2$.

2. Multiply by the number of arrangements of the unique charms within each group to find the total.

Since each charm is uniquely designed, the charms within each group can be arranged among themselves in $5!$, $3!$, and $2!$ ways respectively.
$\text{Total ways} = 2! \times 5! \times 3! \times 2!$
$\text{Total ways} = 2 \times 120 \times 6 \times 2 = 2880$

1. Find the number of ways to arrange the non-small charms in a line

There are $3$ medium and $2$ large charms, giving a total of $5$ non-small charms. Since all charms are uniquely designed, the number of ways to arrange them in a line is:
$5! = 120$

2. Place the small charms into the gaps between the non-small charms and calculate the total ways

The $5$ non-small charms create $6$ available gaps (including the two ends). To ensure no small charms are next to each other, we must place the $5$ uniquely designed small charms into these $6$ gaps.
Number of ways to slot them in $= {^6P_5} = 720$.
Total number of arrangements $= 120 \times 720 = 86400$.

1. Formulate the length equation and identify the constraints for the charms.

Let $s$, $m$, and $l$ be the number of small, medium, and large charms used respectively.
Given the diameters, the total length is $0.5s + m + 2l = 6$.
The question states Kitty has a maximum of 5 small, 3 medium, and 2 large charms, and she uses at least one of each size.
Therefore, the constraints are $1 \le s \le 5$, $1 \le m \le 3$, and $1 \le l \le 2$.

2. Determine all valid combinations of $(s, m, l)$ that satisfy the constraints.

Substitute the possible values of $l$ to find valid $(s, m, l)$ pairs:

If $l = 1$:
$0.5s + m + 2 = 6 \implies s + 2m = 8$
Checking values for $1 \le m \le 3$:

• If $m = 3$, $s = 2$ (Valid)
• If $m = 2$, $s = 4$ (Valid)
• If $m = 1$, $s = 6$ (Invalid, since $s \le 5$)

If $l = 2$:
$0.5s + m + 4 = 6 \implies s + 2m = 4$
Checking values for $1 \le m \le 3$:

• If $m = 1$, $s = 2$ (Valid)
• If $m = 2$, $s = 0$ (Invalid, since $s \ge 1$)

Thus, the valid cases for $(s, m, l)$ are $(4, 2, 1)$, $(2, 3, 1)$, and $(2, 1, 2)$.

3. Calculate the number of ways to choose and arrange the charms for each valid case.

Since all charms are uniquely designed, order matters, and the ends of the chain are distinct (one has a keyring), we use permutations for the selected charms.

Case 1: $(s=4, m=2, l=1)$, total $7$ charms.
Ways $= \binom{5}{4} \times \binom{3}{2} \times \binom{2}{1} \times 7! = 5 \times 3 \times 2 \times 5040 = 151200$

Case 2: $(s=2, m=3, l=1)$, total $6$ charms.
Ways $= \binom{5}{2} \times \binom{3}{3} \times \binom{2}{1} \times 6! = 10 \times 1 \times 2 \times 720 = 14400$

Case 3: $(s=2, m=1, l=2)$, total $5$ charms.
Ways $= \binom{5}{2} \times \binom{3}{1} \times \binom{2}{2} \times 5! = 10 \times 3 \times 1 \times 120 = 3600$

4. Find the total number of different ways.

Total number of ways $= 151200 + 14400 + 3600 = 169200$