twelve people are to travel by three different cars,each of which holds four.Find the number of ways in which the party may be divided if two people refuse to travel in the same car;
---------------------------------------------------------
(1)Assume the antagonists travel in the same car;
Let the cars be a Ford (car number 1), a Chevy (car number 2), and a
Honda (car number 3)
Let the antagonists be John and Henry.
First find the number of ways anybody can ride in any car.
We can choose the four for the Ford in C(12,4) ways.
We can choose the four for the Chevy in C(8,4) ways.
We must choose the four for the Honda in C(4,4)=1 way.
That's C(12,4)·C(8,4)·C(4,4) = 495·70·1 = 34650
From that number we must subtract the number of ways John and
Henry can ride in the same car.
We can choose the car they ride in in C(3,1)=3 ways.
We can choose the other two people to ride with them in C(10,2) ways.
We can choose the four people for the car with the smaller car number
that they don't ride in in C(8,4) ways.
We can choose the four for the car with the larger car number that
they don't ride in in C(4,4)=1 way.
That's C(3,1)·C(10,2)·C(8,4)·C(4,4) = 3·45·70·1 = 9450
So the answer is 34650-9450 = 25200
------------------------------------------
(2)Solve by actually placing the two in different cars.
We can choose the car for John to ride in in C(3,1)=3 ways.
We can choose the car for Henry to ride in in C(2,1)=2 ways.
We can choose the 3 people to ride with John in C(10,3) ways.
[Notice at this point we have 4 people seated in the car that
has John in it and Henry is sitting alone in another car. So
we have 5 people seated so far. That leaves 12-5 or 7 people
left to seat.]
We can choose the 3 people to ride with Henry in C(7,3) ways.
We can choose the 4 remining people to ride in the remaining car
in C(4,4)=1 way.
That's C(3,1)·C(2,1)·C(10,3)·C(7,3)·C(4,4) = 3·2·120·35·1 = 25200
----------------------------------------
finally,compare your answers in (1)and (2).
They are exactly the same, both being 25200 ways.
Edwin
