A gathering of 12 people break into 3 groups of 4 to play 3 different games: carrom, UNO and bridge.

1. Formulate the product of combinations to assign people to the 3 distinct games.

$\text{Number of ways} = {^{12}C_4} \times {^8C_4} \times {^4C_4}$

2. Evaluate the combinations to find the total number of ways.

$= 495 \times 70 \times 1$
$= 34650$

1. Find the total unrestricted number of ways to group the 12 people into 3 distinct games (using the result of part a).

$\text{Total unrestricted ways} = \binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = 495 \times 70 \times 1 = 34650$

2. Calculate the number of ways where Aaron and Sandy are placed in the same group.

Choose $1$ of the $3$ games for them to play together: $\binom{3}{1} = 3$ ways.
Choose $2$ out of the remaining $10$ people to join them: $\binom{10}{2} = 45$ ways.
Group the remaining $8$ people into the other $2$ games: $\binom{8}{4} \times \binom{4}{4} = 70 \times 1 = 70$ ways.

$\text{Ways they are together} = 3 \times 45 \times 70 = 9450$

3. Use the complement method to find the number of ways they are not in the same group.

$\text{Ways they are separated} = 34650 - 9450 = 25200$

1. Determine the total number of ways the 12 people can be grouped such that Aaron and Sandy are not in the same group (denominator).

First, we find the result from part (b) representing our conditional sample space.
Total ways to assign 12 people to 3 specific games = $\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = 495 \times 70 \times 1 = 34650$

Ways Aaron and Sandy are in the same group:

• Choose 1 of the 3 games for them: 3 ways
• Choose 2 more people from the remaining 10: $\binom{10}{2} = 45$ ways
• Distribute the remaining 8 people into the 2 remaining games: $\binom{8}{4} \times \binom{4}{4} = 70$ ways
  Ways they are together = $3 \times 45 \times 70 = 9450$

Ways Aaron and Sandy are in different groups = $34650 - 9450 = 25200$

2. Calculate the number of ways Billy is in Aaron's group while Sandy is in a different group (numerator).

• Choose a game for Aaron and Billy: 3 ways
• Choose 2 more people for their group from the remaining 9 (excluding Sandy): $\binom{9}{2} = 36$ ways
• Choose a game for Sandy from the remaining 2 games: 2 ways
• Choose 3 more people for Sandy's group from the remaining 7: $\binom{7}{3} = 35$ ways
• The last 4 people take the last game: $\binom{4}{4} = 1$ way

Number of ways = $3 \times 36 \times 2 \times 35 \times 1 = 7560$

3. Calculate the conditional probability.

Using the conditional probability formula $P(A|B) = \frac{n(A \cap B)}{n(B)}$:
$\text{P}(\text{Billy with Aaron} \mid \text{Aaron and Sandy separated}) = \frac{7560}{25200}$
$\text{P} = \frac{756}{2520} = \frac{3}{10}$