Problem Solving

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

24

32

48

60

192








i did the hard part- 8*6*4.
But the solution by MGMAT is :


I know in vague terms that this is permutation hence my result needs to be divided by 3!. In other technical words-order is not imp hence divide by 3!. But I almost always err in this distinction. Can someone enlighten using a layman example ?

Hi,
You have 4 colors.
Pick 3 colors from 4 in 4C3 = 4 ways
For each color you can choose either A or B in 2 ways.
So, total number of ways is 4*2*2*2 = 32 ways.

bblast wrote: I know in vague terms that this is permutation hence my result needs to be divided by 3!. In other technical words-order is not imp hence divide by 3!. But I almost always err in this distinction. Can someone enlighten using a layman example ?

Hi,
Lets say the cars are A1,B1,A2,B2,A3,B3,A4,B4
In your method you are picking 1 car(say A1) from 8 is 8C1 = 8 ways
Second car(say B1) from 6 (other colored cars) in 6ways and 3rd(say C1) in 4 ways
Total number of ways is 8*6*4 = 192 ways
In essence you are counting this trio in the following ways
A1,B1,C1
A1,C1,B1
B1,C1,A1
B1,A1,C1
C1,A1,B1
C1,B1,A1
But, in the question we need to only find the combination, so order is not important. So, each combination as it is counted 6 times, in order to get the number of combinations we need to divide by 6(3!).

bblast wrote:The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

24

32

48

60

192

i did the hard part- 8*6*4.
But the solution by MGMAT is :


I know in vague terms that this is permutation hence my result needs to be divided by 3!. In other technical words-order is not imp hence divide by 3!. But I almost always err in this distinction. Can someone enlighten using a layman example ?

I use the slot method:

1. Draw a slot for each choice that must be made.
2. Determine the number of choices for each slot.
3. If order matters -- if we're being asked to count the number of arrangements -- multiply the numbers in each slot. The product is the number of possible arrangements.
4. If order doesn't matter -- if we're being asked to count the number of combinations -- divide the product above by (the number of slots)!. The result is the number of possible combinations.

In the problem above:
We're choosing 3 cars, so we need 3 slots.
Number of choices for the first slot = 8. (We can choose from 8 cars.)
Number of choices for the second slot = 6. (We can choose from the 6 cars that are of a different color from that of car 1)
Number of choices for the third slot = 4. (We can choose from the 4 cars that are of a different color from that of car 1 and that of car 2.)

If the question asked for the number of ways that the cars could be displayed in a row -- if the order of the cars mattered -- we would multiply the numbers above:
8*6*4 = 192 arrangements.
The result is large because more arrangements are possible than are combinations: BRG is a different arrangement from RGB.

The problem above asks us to count the number of combinations, so the product above needs to be divided by (the number of slots)!:
(8*6*4)/3! = (8*6*4)/(3*2*1) = 32.
The result is smaller because fewer combinations are possible than are arrangements: while BRG is a different arrangement from RGB, BRG is the same combination as RGB.
Dividing by (the number of slots)! accounts for all the duplicate combinations.

As you noted, (the number of slots)! = the number of ways to arrange the 3 cars. By dividing by the number of ways that BRG can be arranged, we make sure that we're not overcounting BRG, RBG, etc. as different combinations.

Thank u Frank and than u Mitch, I will try to hammer my brain not to err at this step on the next question I face on P&C.