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? a) 24 b) 32 c) 48 d) 60 e) 192

There are 2 models A and B and 4 colors each. So there are 8 unique combinations. The first car can be selected in 8 ways. Since you selected one color there are three colors left among two models. i.e. 6 cars. The second car can be selected in 6 ways. Since you selected two colors there are two colors left among two models. i.e. 4 cars. The second car can be selected in 6 ways. Total number of different combinations of three cars can the carsons select if all the cars are to be different colors= 8*6*4/3! = 32 Answer is B

niketdoshi123 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? a) 24 b) 32 c) 48 d) 60 e) 192 brainturner did it correctly using FCP. Here's another way: We need three colors out of 4 colors, number of ways of selecting 3 out of 4 = 4C3 = 4 Now for each of the colors, we have the option of either model A or model B. In other words we have 2 selections per color. Then the total number of selections = 4C3 * 2* 2* 2 = 4*8 = 32. Cheers!

Brainturner: I understood how you selected: Car 1 in 8 ways (2 models * 4 colours) Car 2 in 6 ways (2 models * 3 remaining colours) Car 3 in 4 ways (2 models * 2 remaining colours) I also understood how you got the expression: 8*6*4. But why did you divide that by 3! ? Please can you explain?

Hey Kartikshah When we multiplied 8*6*4 we are counting the number of permutations where the order is not important. Since the question asks us to find different combinations we divide by 3! to remove repetitions. We are dividing by 3! because each combination shall comprise of three cars. For example A Red A Black A Green A Red A Green A Black A Green A Black A Red A Green A Red A Black A Black A Red A Green A Black A Green A Red All of these 6 possibilities are just the same with a different order. Hope this explanation helps.