My Math Forum Probability Question

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 January 24th, 2010, 06:16 PM #1 Newbie   Joined: Jan 2010 Posts: 14 Thanks: 0 Probability Question there are 9 people going for a skiing, to travel in skiing lodge there are 3 cars that they can used, the first car can occupy 2 people, the second car can occupy 4 people and the last car can occupy 5 people.. The Question is in how many ways they can travel on skiing lodge using the all cars? Hope You Help me answer this question
January 25th, 2010, 05:47 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Probability Question

Hello, herohenson!

There is no single formula for this problem.
We must do some Listing.

Quote:
 There are 9 people going to a ski lodge. There are 3 cars that they can use. [color=beige]. . [/color]Car A can hold 2 people, [color=beige]. . [/color]Car B can hold 4 perople, [color=beige]. . [/color]Car C can hold 5 people. In how many ways they can travel using the all cars?

I will assume that the order of the people in a car does not matter.
(That is, we don't consider where each person sits, or who is the driver.)

Since all three cars must be used, there are 5 cases to consider.

[color=beige]. . . [/color]$\text{Case} \qquad\qquad \text{CarA} \qquad \text{CarB} \qquad \text{CarC}$

[color=beige]. . . [/color]$\:\qquad 1 \qquad\qquad\qquad\qquad\qquad 1 \qquad\qquad\qquad\qquad\; 3 \qquad\qquad\qquad\qquad 5$

[color=beige]. . . [/color]$\qquad 2 \qquad\qquad\qquad\qquad\qquad 1 \qquad\qquad\qquad\qquad\; 4 \qquad\qquad\qquad\qquad 4$

[color=beige]. . . [/color]$\qquad 3 \qquad\qquad\qquad\qquad\qquad 2 \qquad\qquad\qquad\qquad\; 2 \qquad\qquad\qquad\qquad 5$

[color=beige]. . . [/color]$\qquad 4 \qquad\qquad\qquad\qquad\qquad 2 \qquad\qquad\qquad\qquad\; 3 \qquad\qquad\qquad\qquad 4$

[color=beige]. . . [/color]$\qquad 5 \qquad\qquad\qquad\qquad\qquad 2 \qquad\qquad\qquad\qquad\; 4 \qquad\qquad\qquad\qquad 3$

$\text{In Case 1, there are: }\:\frac{9!}{1!\,3!\,5!} \;=\;504\text{ ways.}$

$\text{In Case 2, there are: }\:\frac{9!}{1!\,4!\,4!} \:=\:630\text{ ways.}$

$\text{In Case 3, there are: }\:\frac{9!}{2!\,2!\,5!} \:=\:756\text{ ways.}$

$\text{In Case 4, there are: }\:\frac{9!}{2!\,3!\,4!} \:=\: 1260\text{ ways}$

$\text{In Case 5, there are: }\:\frac{9!}{2!\,4!\,3!} \:=\:1260\text{ ways.}$

$\text{Therefore, there are: }\:504\,+\,630\,+\,756\,+\,1260\,+\,1260 \;=\;4410\text{ ways.}$

 January 25th, 2010, 07:41 PM #3 Newbie   Joined: Jan 2010 Posts: 14 Thanks: 0 Re: Probability Question sir thank you i got the idea..

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