Permutations & Combinations: Groups of 12; restriction — ACJC 2025 H2 Math Prelim Paper 2

Part (a)

Select 4 from 12 for bridge: \(\binom{12}{4}\). Select 4 from remaining 8 for carrom: \(\binom{8}{4}\). Remaining 4 play mahjong: \(\binom{4}{4}\).

By the multiplication principle: \[ \binom{12}{4}\times\binom{8}{4}\times\binom{4}{4} = 495 \times 70 \times 1 = 34650. \]

Part (b)

Method 1: By complementation

Total ways with Aaron and Sandy in the same group: place them together in one of 3 groups (\(\times 3\)), choose 2 from 10 to complete that group (\(\binom{10}{2}\)), then assign remaining 8 (\(\binom{8}{4}\)): \[ 3\times\binom{10}{2}\times\binom{8}{4}\times\binom{4}{4} = 3 \times 45 \times 70 \times 1 = 9450. \] Number with Aaron and Sandy in different groups: \[ 34650 - 9450 = 25200. \]

Method 2: Direct

Assign Aaron to any of 3 groups (\(\times 3\)), Sandy to one of the other 2 (\(\times 2\)), complete Aaron's group from 10 (\(\binom{10}{3}\)), complete Sandy's group from 7 (\(\binom{7}{3}\)): \[ 3 \times 2 \times \binom{10}{3} \times \binom{7}{3} \times \binom{4}{4} = 3\times 2\times 120\times 35\times 1 = 25200. \]

Part (c)

Let \(A = \{\text{Aaron and Billy in same group}\}\), \(B = \{\text{Aaron and Sandy in different groups}\}\).

Place Aaron and Billy together (3 choices of group), Sandy in one of the other 2 groups, choose 2 from 9 remaining to complete Aaron/Billy's group, choose 3 from 7 for Sandy's group: \[ n(A\cap B) = 3 \times 2 \times \binom{9}{2} \times \binom{7}{3} \times 1 = 3\times 2\times 36\times 35 = 7560. \]

From part (b): \(n(B) = 25200\).