There are total 4410 ways possible to transport the 9 people to the ski lodge, using all cars.
It is given that the nine people are going on a skiing trip in 3 cars that hold 2, 4, and 5 passengers, respectively.
It is required to find them in how many ways it is possible to transport the 9 people to the ski lodge, using all cars
A permutation can be defined as the number of ways a set can be arranged and the arrangements of the set matter.
We have nine people if all the nine people are distinguishable.
If the car that holds 2 passengers holds one passenger then other cars have to hold 8 passengers, for this we have two options:
1) One car can hold 4 (capacity is 4) passengers and another car hold 4 (Capacity is 5) passengers.
2) One car can hold 3 (capacity is 4) passengers and another car hold 5 (Capacity is 5) passengers.
So from the permutation formula: [tex]\rm _{n}^{}\textrm{P}_{r} = \frac{n!}{(n-r)!}[/tex]
[tex]\rm(4,4):\frac{8!}{(8-4)! \ (8-4)!}\ \Rightarrow \frac{8!}{4! \ 4!}\ \Rightarrow 56 \ ways\\\rm \\ (3,5):\frac{8!}{(8-3)! \ (8-5)!} \Rightarrow \frac{8!}{5! \ 3!} \Rightarrow 70 \ ways[/tex]
Hence, there are 56+70=126 ways to seat the 8 passengers.
For the 9 people = 9×126 = 1134 ways.
If the car holds two passengers then other cars have to hold 7 passengers, then:
[tex]\rm \frac{9!}{2!} \Rightarrow 36[/tex] ways for the two passengers.
And for the other 7 passengers there are we have three options:
Similarly, using the permutation formula, we get
[tex]\rm (2,5) : \frac{7}{2! \ 5!} \Rightarrow 21 ways\\\\\rm (3,4) : \frac{7}{3! \ 4!} \Rightarrow 35 ways\\\\\rm (4,3) : \frac{7}{4! \ 3!} \Rightarrow 35 ways\\[/tex]
So, there are 21+35+35= 91 ways to seat the 7 people.
Hence, there are 36×91 = 3276 ways.
Total ways = 1134 + 3276 = 4410 ways.
Thus, There are total 4410 ways possible to transport the 9 people to the ski lodge, using all cars.
Learn more about permutation here:
brainly.com/question/1216161
