Total types of cars

Model A - Blue
Model A - Black
Model A - Red
Model A - Green
Model B - Blue
Model B - Black
Model B - Red
Model B - Green

Number of ways of selecting the first car : 8(let us assume that the family selected a Red Car, then the family cannot select another Red car. So, Only 6 cars are left for the next seleionction)

Number of ways of selecting the Second car : 6(let us assume that the family selected a Blue Car, then the family cannot select another Blue car. So, Only 4 cars are left for the next selection)

Number of ways of selecting the Third car : 4

Total number of ways = c

But a car combination
Red-Green-Blue is same as
Red-Blue-Green, which is same as
Green-Blue-Red, which is same as
Green-Red-Blue, which is same as
Blue-Red-Green, which is same as
Blue-Green-Red.

So, same combination repeats 6 times, so the number combinations of three cars can the Carsons select if all the cars are to be different colors? = 8*6*4/6 = 32

thanks

Here Model is fewer than the number of Cars to be purchased.

Let's break down the problem

[color=blue]CASE I:[/color]

All 3 colors are of one model

So Combination is 4C3= 4 and there are 2 models so 4*2 =8

CASE II:

One cars is of color of one model and the remaining two are of different colors of the 2nd model = 4C1*3C2=12. This also will happen for 2 models. So total 2*12=24


Combining cas I and II . Total Combinations = 8+24 = 32