[Shogun]

Three of the students (Sam, Bill, and Steve) own cars and will provide transportation to the game. Sam?s car has room for three passengers, while the other cars owned by Bill and Steve each have room for four passengers. In how many different ways can the passengers be loaded into the cars?

[ebaines]

I think you left out an important bit of information - how many passsengers require transportation?

Also - please try to solve this yourself and post what you think the answer is and why. We'll check it for you and help you out if you get stuck. But please show us you tried.

[Shogun]

Sorry, there is 14 total people. Three are drivers, so that leaves that is 11 additional students needing rides.

Truthfully I am 100% stuck on the problem, but I am thinking something similar to this type of math:

11!*10!*9!*8!*7!*6!*5!*4!*3!*2!

But that number seems a bit high for that to make sense to me so there has to be another method to finding this.

[Shogun]

Sam - 1 ,2 ,3
Bill - 4 ,5 ,6 ,7
Steve = 8 ,9 ,10 ,11

That could be the 1st possible setup. I understand how to solve it if they were all going together in one car, but not how to go about it if they are split between 3 different cars.

[ebaines]

Since there are 11 passengers and room for 11 passengers between the three cars that means all cars are filled. One way to solve this is to think how many ways can 11 passengers be arranged into 11 seats? That would be 11 x 10 x 9 x... x 1 = 11! Now consider that for each car it doesn't matter which student sits where in that car - in other words if Sam has 3 passengers named A, B and C it doesn't matter whether they are seated in his car in order ABC, ACB, BAC, BCA, CAB or CBA. Hence to eliminate all these duplicates you must divide by 3! Same thinking applies to the 4 people in Bill's car and the 4 in Steve's. Can you take it from here?

[Shogun]

So would it be math something like this:

(11!) = 165
(11-3)!3!

Then for the others:

(11!) = 330
(11-4)!4!

Then multiply the numbers together to get 330+330+165 = 825

That number looks a lot more reasonable than my first attempt at it, but seems a bit low for 11 people riding in 3 separate cars. Thanks for the help thus far ebaines, I was legitimately confused about how to solve this problem.

[ebaines]

Not quite..

First of all, you have written 11! = 165, and also 11! = 330, neither of which is correct. 11! Is equal to 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 which is a huge number. It accounts for every possible arrangement of all 11 passengers into the 11 seats in the 3 cars. You need to divide 11! By the number of ways that 3 students can be arranged in one car (which is 3!), 4 students can be arranged in the second car (4!), and 4 students can be arranged in the 3rd car (4!).

Here's an example of a similar problem - you can use the same technique for your problem: How many ways can 8 people be assigned to two classrooms, given that 3 must be assigned to one class and 5 to the other? There are 8! Ways to assign the 8 students to individual seats in the classrooms, but for the first class it doesn't matter what order the 3 students are placed, and in the second class it also doesn't matter what order; so the answer is:



Can you apply the same reasoning to your problem?

[Daniel01]

A summer camp wants to hire counselors and aides to fill its staffing needs at a minimum cost. The average monthly salary of a counselor is $2400 and the average monthly salary of an aide is $1100. The camp can accommodate up to 45 staff members and needs at least 30 to run properly. They must have at least 10 aides, and may have up to 3 aides for every 2 couselors. How many counselors and how many aides should the camp hire to minimize cost?