Suppose 14 students have tickets for a concert a Three stude

Suppose 14 students have tickets for a concert.

(a) Three students (Bob, Jim, and Tom) own cars and will provide transportation to the concert. Bob’s car has room for three passengers (nondrivers), while the cars owned by Jim and Tom each has room for four passengers. In how many different ways can the 11 passengers be loaded into the cars.

(b) At the convert hall the students are seated together in a row. If they take their seats in random order, find the probability that the three students who drove their cars have adjoining seats.

Solution

Total students have ticket for a concert = 14

There are three students Bob,Jim and Tom has own cars.

Bob\'s car has room for three passengers while Jim and Tom\'s car has room for four passengers.

a) In how many different ways can the 11 passengers be loaded into the cars.

Here we use combination concept-

Bob\'s car has capacity of 3 passengers it means out of 14 the 3 passengers are seated in the car in (14 C 3) ways.Jim\'s car has capacity of 4 passengers it means out of 14 the 4 passengers are seated in the car in (14 C 4) ways. Tom\'s car has capacity of 4 passengers it means out of 14 the 4 passengers are seated in the car in (14 C 4) ways.

= (14 C 3)*(14 C 4)*(14 C 4) (Here C is used for combinations)

Formula- (n C r) = n! / r! * (n-r)!

=(14! / 3!*11!) * (14! / 4!*10!) * (14! / 4!*10!)

n! = n *(n-1) !

=364728364 different ways 11 passengers be loaded into the cars.

And if no passenger will be repeated then

( 14 C 3)*(11 C 4)*(7 C 4) = 4204200 different ways passengers be loaded into the cars.

b) At the convert hall the students are seated together in a row. If they take their seats in a random order.

To  find the probability that the three students who drove their cars have adjoining seats.

It means Bob , Jim and Tom take the adjoining seats.

P ( the three students who drove their cars have adjoining seats )

There are 11! ways to place 11 students are seated together in a row. There are 9! ways to arrange a group of Bob,Jim and Tom and 8 other students. Among the three students (in a group), they can arrange themselves in 3! different ways.

So the probability that they will be seated together (adjoninig seats) is,

( 9! * 3! ) / 11! = 0.054545