凯莉（Kelly）有一盒标有从 A 到 Z 26 个字母的球形珠子，所有珠子的大小都相同。对于每个字母，恰好有 3 颗不同颜色的珠子：红色、蓝色和金色。凯莉计划通过将这些标有字母的珠子按特定顺序串起来制作一个钥匙扣。图 1 展示了一个带有 3 颗标有字母的珠子的钥匙扣可能的设计。

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:
$\binom{45}{2} = \frac{45 \times 44}{2} = 990$

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