Kelly has a box of spherical beads labelled with the 26 letters from A to Z, all of which have the same size. For each letter, there are exactly 3 beads of different colours, red, blue and gold. By stringing the labelled beads together in a specific order, Kelly plans to make a keychain. A possible design of a keychain with 3 labelled beads is shown in Fig. 1.

1. Determine the number of ways to choose the colour for the beads.

Since all 6 beads must be of the same colour, and there are 3 possible colours (red, blue, and gold), there are 3 ways to choose the colour.

2. Calculate the number of ways to arrange 6 distinct letters out of the 26 available letters for the chosen colour.

For a specific colour, there is exactly 1 bead of each letter, giving 26 distinct beads. The number of ways to choose and arrange 6 distinct beads in a specific order is:
$^{26}P_6 = \frac{26!}{(26-6)!} = 26 \times 25 \times 24 \times 23 \times 22 \times 21 = 165\,765\,600$

3. Calculate the total number of arrangements.

Total number of ways $= 3 \times 165\,765\,600 = 497\,296\,800$

1. Find the number of ways to select 3 distinct letters.

We need exactly 3 distinct letters for the 6 beads. The number of ways to choose 3 letters out of the 26 available is:
$\binom{26}{3} = 2600$

2. Find the number of ways to select the 6 beads for a given set of 3 letters.

Each letter has exactly 3 available beads (Red, Blue, Gold). To have exactly 3 distinct letters among 6 beads, the distribution of beads per letter must be partitioned into either $3, 2, 1$ or $2, 2, 2$.

Case 1: Bead distribution is 3, 2, 1
Ways to assign these bead counts to the 3 chosen letters $= 3! = 6$
Ways to choose the beads for these letters $= \binom{3}{3} \times \binom{3}{2} \times \binom{3}{1} = 1 \times 3 \times 3 = 9$
Total ways for this case $= 6 \times 9 = 54$

Case 2: Bead distribution is 2, 2, 2
Ways to choose 2 beads for each of the 3 letters $= \binom{3}{2} \times \binom{3}{2} \times \binom{3}{2} = 3 \times 3 \times 3 = 27$

Total ways to select the 6 beads $= 54 + 27 = 81$

3. Calculate the total number of arrangements.

Since every bead is unique (having a distinct letter and colour combination), the chosen 6 beads are all distinct. The number of ways to arrange them in a sequence is $6!$.

Total number of arrangements:
$\text{Total ways} = 2600 \times 81 \times 6!$
$\text{Total ways} = 210,600 \times 720 = 151,632,000$

1. Express the total number of beads and the number of target beads in terms of $c$.

$\text{Total number of beads} = (c + 6 + 2) + (2 + 5 + 5) + (6 + 3 + 2) + (5 + 2 + 3) = c + 41$

The star-shaped beads that are not gold are the red stars and blue stars.
$\text{Number of star-shaped, non-gold beads} = c + 2$

2. Formulate an equation based on the probability of drawing two star-shaped, non-gold beads without replacement.

$\text{P(both are star-shaped and not gold)} = \left(\frac{c+2}{c+41}\right) \left(\frac{c+1}{c+40}\right)$

Given that this probability is $\frac{1}{66}$:
$\frac{(c+2)(c+1)}{(c+41)(c+40)} = \frac{1}{66}$

3. Solve the resulting quadratic equation to find $c$.

$66(c+2)(c+1) = (c+41)(c+40)$
$66(c^2 + 3c + 2) = c^2 + 81c + 1640$
$66c^2 + 198c + 132 = c^2 + 81c + 1640$
$65c^2 + 117c - 1508 = 0$

Dividing by 13:
$5c^2 + 9c - 116 = 0$
$(c - 4)(5c + 29) = 0$

$c = 4 \quad \text{or} \quad c = -5.8$

Since $c$ represents a count of objects, it must be a positive integer. Therefore, $c = 4$.

1. Find the value of $c$ using the information given in part (b)(i).

Let the total number of beads be $N$. Summing all beads in the table:
$N = (c + 6 + 2) + 12 + 11 + 10 = c + 41$
The number of star-shaped beads that are not gold (i.e., red or blue) is $c + 2$.
The probability of choosing 2 such beads is $\frac{1}{66}$:
$\frac{c+2}{c+41} \times \frac{c+1}{c+40} = \frac{1}{66}$
$66(c^2 + 3c + 2) = (c+41)(c+40)$
$66c^2 + 198c + 132 = c^2 + 81c + 1640$
$65c^2 + 117c - 1508 = 0$
Dividing by 13:
$5c^2 + 9c - 116 = 0$
$(5c + 29)(c - 4) = 0$
Since $c$ must be positive, $c = 4$.
Thus, the total number of beads in the box is $4 + 41 = 45$.

2. Group the 45 beads based on their shape (Flower or Not Flower) and colour (Red or Not Red).

Using the table and $c=4$, we categorise the beads:

• Flower & Red: $6$
• Flower & Not Red: $3 \text{ (Gold)} + 2 \text{ (Blue)} = 5$
• Not Flower & Red: $4 \text{ (Star)} + 2 \text{ (Butterfly)} + 5 \text{ (Heart)} = 11$
• Not Flower & Not Red: Total beads - $(6 + 5 + 11) = 45 - 22 = 23$

3. Calculate the number of valid ways to choose exactly one red bead and at least one flower bead.

We need exactly 1 Red bead and 1 Not Red bead. For the pair to contain at least 1 Flower, the valid combinations are:

1. (Flower & Red) and (Flower & Not Red):
   $6 \times 5 = 30 \text{ ways}$
2. (Flower & Red) and (Not Flower & Not Red):
   $6 \times 23 = 138 \text{ ways}$
3. (Not Flower & Red) and (Flower & Not Red):
   $11 \times 5 = 55 \text{ ways}$

Total valid ways $= 30 + 138 + 55 = 223$.

4. Calculate the required probability.

The total number of ways to choose any 2 beads from the 45 beads is:
$\binom{45}{2} = \frac{45 \times 44}{2} = 990$

$P(\text{exactly 1 red, at least 1 flower}) = \frac{223}{990}$