Some MRT trains have cars featuring 12 seats each. Passengers take the trains to travel from station to station each day. Seats located nearest to the train doors are labelled as priority seats reserved for vulnerable commuters, namely those who are pregnant, elderly, with infants, or the differently-abled. Anyone can sit at the priority seats if there are sufficient seats for all vulnerable commuters in the car. For one trip, 12 passengers, of which 4 are vulnerable commuters, got onto Car B at the first station.

1. Determine the number of ways to seat the 4 vulnerable commuters in the 4 priority seats.

There are exactly $4$ priority seats in Car B. The $4$ vulnerable commuters must sit in these seats.
Number of ways to seat them = $4!$

2. Determine the number of ways to seat the remaining passengers and calculate the total number of arrangements.

There are $8$ remaining passengers to be seated in the $8$ ordinary seats.
Number of ways to seat them = $8!$

Total number of ways = $4! \times 8! = 24 \times 40320 = 967680$

1. Calculate the total number of ways to arrange all 12 passengers without any restrictions.

$\text{Total unrestricted ways} = 12!$

2. Calculate the number of ways to arrange the passengers such that the 2 particular girls are seated together, by treating them as a single grouped unit.

There are $12 - 2 + 1 = 11$ units to arrange (10 individual passengers + 1 group of 2 girls).
$\text{Ways the girls sit together} = 11! \times 2!$

3. Subtract the number of ways they sit together from the total unrestricted ways to find the number of arrangements where they are not beside each other.

$\text{Ways they do not sit together} = 12! - (11! \times 2!)$
$= 479\,001\,600 - 79\,833\,600$
$= 399\,168\,000$

1. Find the number of ways to select and arrange 4 out of the 5 girls to sit together.

Choose 4 girls to form a group and arrange them within the group:
$\text{Number of ways} = \binom{5}{4} \times 4! = 5 \times 24 = 120$

2. Find the number of ways to arrange the remaining 7 passengers.

There are $12 - 5 = 7$ other passengers.
$\text{Number of ways to arrange them} = 7! = 5040$

3. Use the gap method to place the group of 4 girls and the remaining 1 girl so they are separated.

The 7 other passengers create $8$ available spaces (including the ends of the row). To ensure exactly 4 girls are together, the group of 4 girls and the single remaining girl must be placed in different spaces.
$\text{Number of ways to assign spaces} = {}^{8}P_2 = 8 \times 7 = 56$

4. Calculate the total number of valid seating arrangements.

Multiply the number of ways from the independent choices:
$\text{Total ways} = 120 \times 5040 \times 56 = 33,868,800$