In a strategy game called Colour Clash, players draw coloured orbs from a magical pouch to determine their elemental powers. The pouch contains:

• 5 red orbs (Level 1 to 5),
• 4 blue orbs (Level 1 to 4),
• 3 green orbs (Level 1 to 3).

Orbs of the same colour are non-identical with each same-coloured orb representing a different level. A player randomly draws four orbs without replacement at the start of the game. Given that the order in which the orb is drawn is not relevant, find the number of ways this can be done such that

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

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