Nine IU South Bend students are going on a trip to Indianapolis for a Clots game. Go Colts!(a) They will seat in a row. In how many ways can they seat if Josh, Mark and Mary will sit together?(b) They will travel with three cars that hold 2, 4, and 5 people, respectively. All of them can drive. Find the numberof possible ways to allocate 9 people to 3 cars that each car has at least one person (each car has to have a driver)(i) if the allocations are to the cars only (disregard the seats in the car). (ii) if the allocations are to the seats.

a) To find the number of ways the nine IU South Bend students can be seated if Josh, Mark, and Mary must sit together, we can treat these three students as a single group. This means we have 7 entities to arrange: the group of Josh, Mark, and Mary, and the other 6 students.Now, within the group of Josh, Mark, and Mary, they can be arranged among themselves in 3 ! (3 factorial) ways.Additionally, you have 7 entities in total to arrange, which can be done in 7 ! ways.So, the total number of ways they can be seated is 3 ! (for the arrangement within the group) multiplied by 7 ! (for the arrangement of all 7 entities):3 ! * 7 !=6 * 5040=30,240 ways.Therefore, there are 30,240 ways to seat the students if Josh, Mark, and Mary must sit together.b) (i) To find the number of possible ways to allocate 9 people to 3 cars where each car must have at least one person, you can use a partitioning approach. You can think of this as distributing 9 people into 3 groups (the cars) with the constraint that each group has at least one person.You can use stars and bars to visualize this. Imagine 9 stars representing the 9 people, and you want to place 2 bars to divide them into 3 groups (the cars). Each group must have at least one star (the driver). This problem is equivalent to finding the number of ways to arrange these 9 stars and 2 bars.The number of ways to arrange these symbols (stars and bars) is given by the binomial coefficient " " because you are choosing 2 positions out of 11 total positions for the bars.C(11,2)=55Math input errorMath input errorSo, there are 55 possible ways to allocate the 9 people to the 3 cars if you only consider the allocations to the cars, disregarding the seats in the car.(ii) Now, if you want to consider the allocations to the seats within the cars, you can calculate it as follows:Car 1 ( 2 seats): There are 9 choices for the first person (driver) and 8 choices for the second person.Car 2 ( 4 seats): After the first car is allocated, there are 7 people left. The driver for the second car has 7 choices, and the remaining 3 passengers can be arranged in C(7,3) ways.Car 3 ( 5 seats): After the first two cars are allocated, there are 2 people left. The driver for the third car has 2 choices, and the remaining passenger can occupy any of the 4 remaining seats.Now, you can multiply these choices together to find the total number of ways:Total ways =(9 * 8) *(7 * C(7,3)) *(2 * 4)=9,072 ways.So, there are 9,072 possible ways to allocate the 9 people to the seats within the 3 cars, considering both the allocations to the cars and the seats.a) Therefore, there are 30,240 ways to seat the students if Josh, Mark, and Mary must sit together.b) i) So, there are 55 possible ways to allocate the 9 people to the 3 cars if you only consider the allocations to the cars, disregarding the seats in the car.ii) So, there are 9,072 possible ways to allocate the 9 people to the seats within the 3 cars, considering both the allocations to the cars and the seats.