在一款名为 Colour Clash 的策略游戏中，玩家从一个魔法袋中抽取彩色能量球来确定他们的元素力量。袋中装有：

• 5 个红色球（1 级至 5 级），
• 4 个蓝色球（1 级至 4 级），
• 3 个绿色球（1 级至 3 级）。

相同颜色的球是不相同的，每个相同颜色的球代表不同的等级。 在游戏开始时，玩家随机抽取四个球且不放回。已知抽取的顺序不重要，求在满足下列条件时的抽取方法数：

1. Calculate the total number of ways to select 4 orbs from all available distinct orbs

Total number of distinct orbs $= 5 + 4 + 3 = 12$

Number of ways $= \binom{12}{4} = 495$

1. Identify the complement event and calculate the number of ways all 4 orbs are the same colour.

The complement of 'at least two colours present' is 'all orbs are of the same colour'.
This can only happen if all 4 orbs are red or all 4 are blue (since there are only 3 green orbs).
Number of ways for the same colour = $^{5}C_4 + ^{4}C_4$
$= 5 + 1 = 6$

2. Subtract the complement from the total number of unrestricted ways to find the final answer.

First, find the total unrestricted ways to choose 4 orbs from 12 (from part i):
Total ways = $^{12}C_4 = 495$

Number of ways with at least two colours = Total ways $-$ Ways with only one colour
$= 495 - 6 = 489$

1. Identify the possible mutually exclusive colour combinations to draw 4 orbs with at least 1 of each colour

Since there are 4 orbs drawn from 3 different colours and we need at least 1 of each colour, the only possible colour distributions are:

• Case 1: 2 Red, 1 Blue, 1 Green
• Case 2: 1 Red, 2 Blue, 1 Green
• Case 3: 1 Red, 1 Blue, 2 Green

2. Calculate the number of ways for each case using the multiplication principle and combinations

Case 1 ($2$ R, $1$ B, $1$ G): ${}^5C_2 \times {}^4C_1 \times {}^3C_1 = 10 \times 4 \times 3 = 120$
Case 2 ($1$ R, $2$ B, $1$ G): ${}^5C_1 \times {}^4C_2 \times {}^3C_1 = 5 \times 6 \times 3 = 90$
Case 3 ($1$ R, $1$ B, $2$ G): ${}^5C_1 \times {}^4C_1 \times {}^3C_2 = 5 \times 4 \times 3 = 60$

3. Sum the number of ways from all cases using the addition principle

$\text{Total number of ways} = 120 + 90 + 60 = 270$