skiing trip probobility

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steller

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Nine people are going on a skiing trip in 3 cars that hold 2, 4, 5 passengers respectively. In how many ways is it possible to transport the 9 people to the ski lodge, using all cars?

I have some similar problems in my book.

Would i be correct by:

\(\displaystyle \frac{9!}{2! 4! 5!} = 63 \)

I found the problem online and people say 5 ways but they dont use statistics.
Thanks
 
steller said:
Nine people are going on a skiing trip in 3 cars that hold 2, 4, 5 passengers respectively. In how many ways is it possible to transport the 9 people to the ski lodge, using all cars?

I have some similar problems in my book.

Would i be correct by:

\(\displaystyle \frac{9!}{2! 4! 5!} = 63 \)

I found the problem online and people say 5 ways but they dont use statistics.
Thanks
Click to expand...
Every car has a driver, so the question is how to transport the remaining six people. The number of seats available (not counting drivers) is 8, so we can ask where the two empty seats are. They could both be in the largest car, OR they could both be in the middle car, OR they could be one each in any two of the three cars. I count a total of five ways.
 
DrPhil said:
Every car has a driver, so the question is how to transport the remaining six people. The number of seats available (not counting drivers) is 8, so we can ask where the two empty seats are. They could both be in the largest car, OR they could both be in the middle car, OR they could be one each in any two of the three cars. I count a total of five ways.
Click to expand...
On the other hand, if we eliminate the drivers,

6!/(1! 3! 4!) = 5
 
I'm glad that the good doctor responded before I did. I was thinking of all possibilities for distributing nine individuals among the cars. Like, 81 different ways, just considering the two-seater ... :oops:
 
steller said:
How did you come up with 1! 3! 4! ?
Click to expand...
The available seats for 6 non-drivers in the three cars. Actually, I just copied what you had done, but with drivers' seats all occupied to begin with. Is it just by accident that it gives 5?
 
Hello, steller

Note that "drivers" are not considred.
Nine people are going on a skiing trip in 3 cars that hold 2, 4, 5 passengers respectively.
In how many ways is it possible to transport the 9 people to the ski lodge, using all cars?
Click to expand...
I will assume that the nine people are distinguishable.

Let's name the cars.
. . \(\displaystyle \begin{array}{c}\text{Car }B\text{ holds 2 people.} \\ \text{Car }D\text{ holds 4 people.} \\ \text{Car }E\text{ holds 5 people.}\end{array}\)


Car \(\displaystyle B\) can hold 1 or 2 people.


[1] Suppose car \(\displaystyle B\) has one person.
. . .There are \(\displaystyle 9\) choices for that one person.

. . .Then the other 8 people are to ride in cars \(\displaystyle D\) and \(\displaystyle E.\)
. . . . .There are two scenarios.
. . .\(\displaystyle (D,E) = (3,5)\!:\;{8\choose3,5} = 56\) ways.
. . .\(\displaystyle (D,E) = (4,4)\!:\;{8\choose4,4} = 70\) ways.
. . . . .So, there are: \(\displaystyle 56 + 70 \,=\,126\) ways to seat the 8 people.

. . .Hence, there are: \(\displaystyle 9\cdot126 \,=\,1134\) ways.


[2] Suppose car \(\displaystyle B\) has two people.
. . .There are: \(\displaystyle {9\choose2} \,=\,36\) choices for the two people.

. . .The other 7 people are to ride in cars \(\displaystyle D\) and \(\displaystyle E.\)
. . .There are three scenarios.
. . .\(\displaystyle (D,E) \,=\,(2,5)\!:\;{7\choose2,5} \,=\,21\) ways.
. . .\(\displaystyle (D,E) \,=\,(3,4)\!:\;{7\choose3,4} \,=\,35\) ways.
. . .\(\displaystyle (D,E) \,=\,(4,3)\!:\;{7\choose4,3} \,=\,35\) ways.
. . . . .So, there are: \(\displaystyle 21+35+35 \:=\:91\) ways to seat the 7 people.

. . .Hence, there are: \(\displaystyle 36\cdot 91 \:=\:3276\) ways.


Therefore, there are: \(\displaystyle 1134 + 3276 \:=\:4410\) ways.
 
steller said:
I am confused. Are there 5 ways or 4410 ways? =(
Click to expand...

There are only five! If we only count content.

This is a standard counting question using a generating function.

Look at this webpage.
Scroll down to "alternate form" to see the term \(\displaystyle 5x^9\).
That tells us that there are five to put nine people into containers that hold two, four and five resp. with with no container empty.

Again,​ that is only if we read the question only as content and not order with no car empty.

P.S. Note that if we don't require the use of all three cars, the answer is six.
 
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