Permutations & Combinations: Colour orbs; at-least — JPJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
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 \(3\)).
A player randomly draws four orbs without replacement at the start of the game. Given that the order in which the orbs are drawn is not relevant, find the number of ways this can be done such that
- (i) there are no restrictions,
- (ii) there are at least two colours present,
- (iii) at least one orb of each colour is drawn.
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(i) Total orbs \(= 5+4+3 = 12\). Number of ways to choose any 4: \[ \binom{12}{4} = 495. \]
(ii) “At least two colours” = total \(-\) (all one colour). The only colours with at least 4 orbs are red (\(5\)) and blue (\(4\)): \[ 495 - \binom{5}{4} - \binom{4}{4} = 495 - 5 - 1 = 489. \]
(iii) “At least one of each colour” with 4 orbs from \(\{5R,4B,3G\}\) — split by which colour repeats:
| Case 1: 2R, 1B, 1G | \(\quad\) | \(\binom{5}{2}\binom{4}{1}\binom{3}{1} = 10\cdot 4\cdot 3 = 120\) |
|---|---|---|
| Case 2: 1R, 2B, 1G | \(\binom{5}{1}\binom{4}{2}\binom{3}{1} = 5\cdot 6\cdot 3 = 90\) | |
| Case 3: 1R, 1B, 2G | \(\binom{5}{1}\binom{4}{1}\binom{3}{2} = 5\cdot 4\cdot 3 = 60\) |
Total \(= 120 + 90 + 60 = 270\).