The Carson family will purchase three used cars. There are

[shaw.s.li]

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

I'm having just a little difficulty understanding this problem. I know to use the slot method and that there are a total of 8 different model cars (2*4).

I also understand the slot method is 8 * 6* 4. What I don't understand is why I need to divide by 3!.

The explanation says order doesn't matter, but I didn't think about that in when I was reading the problem. Can someone help me know when I should or shouldn't be dividing by the number of choice I'm making in such a problem?

[jnelson0612]

Hi shaw,
Great question!

Let's review what we know: there are two models of cars, A and B, and four colors of blue, black, red, and green. So our possible cars are:

A-blue
A-black
A-red
A-green
B-blue
B-black
B-red
B-green

As you stated, the first step would be to multiply the possibilities using the slot method. I have three slots for the three cars. In the first slot, I have 8 possibilities. Once that car is selected, I then have six possibilities (all the remaining cars that are not the first car's color). Once the second car is selected, I have four possibilities for the third slot (the four remaining cars that are not the color of either of the first two cars).

By using the slot method, I am calculating the number of combinations of a particular order. For example, I could be saying for this problem that I choose A-red, B-black, then B-green, or B-blue, A-black, and B-red. When I choose one first, then another, then another, that will give you the number of combinations of a particular order. By using this method, you assume that the order in which you pick the cars DOES matter.

In fact, order does not matter in this problem. We just need three cars of different colors; we don't care if we pick a particular car first. We can adjust for the order not being relevant in this type of problem by dividing by the factorial of the number of items on top. In this case, we are picking 3 cars, so we divide by 3!.

To read more about this, read Ron's excellent explanation of a similar problem in this thread: a-commitee-of-3-people-is-to-be-chosen-t9945.html

[shaw.s.li]

Thanks Jamie. That explanation helps. I still have another question that I want you to help me correct my thinking.

I understand that if I don't divide by 3!, I'm picking number of combinations of a particular order, such as the example you game of A-red, B-black, and B-green.

However, I would think that if I picked by this particular order, the actual combinations should be less than if I divided by 3!.

In other words, wouldn't the particular order for A-red, B-black, and B-green be only 1 since there is only one way to have this particular order? That's why when I think about dividing this by 3!, I have a hard time thinking this way because then the number of combinations is even smaller.

I know this thinking isn't right because the question isn't asking how many combinations can I get exactly A-red, B-black, and B-green in this order. I know there are other combinations because you use the word "or" even in your example. Can you help me clarify my thinking?

[jnelson0612]

Hi shaw,
Sorry if my earlier explanation was confusing.

When you use the slot method, you are coming up with the total of all possible orders and are assuming that order matters. For example, let's say that we have four people, ABCD, who are going to sit in three chairs. How many possible arrangements (orders) can we have of the three people?
4*3*2=24 There are 24 possible ways to seat the people.

Now, let's say we just need to choose 3 people out of 4 to sit in the seats and we don't care about order. We now use
4!
3! 1!
which is 4. There are 4 possible groups of 3 that can sit in the seats.

Notice that we can get 4 by taking the result of the slot method, 24, and dividing by the number of slots factorial, 3!. 24/3! is 24/6 or 4.

Hope this helps!

[shaw.s.li]

Thanks, that really helps

[jnelson0612]

Great! :-)

[irvingpjolly]

But the question specifically asks for "combinations", not "permutations". The latter implies order is relevant, the former implies that it is not relevant. In the link provided to the "similar question", the wording is "different committees", which does not explicate whether or not the order is relevant.

You can argue that there is a common sense element to whether or not the order is relevant, but I would bet that if you surveyed test takers, there would not be complete agreement on order relevance for a given question.

I think questions such as these should include explicit wording about the relevance of the order.

- Larry S.

[satyaking]

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

I'm having just a little difficulty understanding this problem. I know to use the slot method and that there are a total of 8 different model cars (2*4).

I also understand the slot method is 8 * 6* 4. What I don't understand is why I need to divide by 3!.

The explanation says order doesn't matter, but I didn't think about that in when I was reading the problem. Can someone help me know when I should or shouldn't be dividing by the number of choice I'm making in such a problem?

alternative approach:

no. of groups not having any two cars of same colour=total no. of possible groups of 3 cars -no.of groups of 3 cars having any two cars of same colour

no.of groups of 3 cars having any two cars of same colour
=6 . 4= 24
scenario1: black_ black_ (any one colour from rest of 6 cars)
= 6 possible ways(order does not matter in group selection)

similarly,all total 4 scenarios for black ,blue ,green ,and red colour respectively,leading to 6.4 i.e 24 ways of selection.

no. of groups not having any two cars of same colour=total no. of possible groups of 3 cars -no.of groups of 3 cars having any two cars of same colour
=8c3 -24
=56-24
=32
hope it helps

[jnelson0612]

Thanks satya--it's always interesting to see different approaches.

[bansalm]

Alternatively, select the 3 different colors first - in 4C3 ways
Then for each selected color, there are 2 choices of models.
Hence, total choices will be 4*2*2*2=32

hope this helps

[tim]

cool; thanks..

[coach]

4!*2!=4*3*2*2=48

[tim]

it does indeed..

[rooproyee]

Wow! Its really very nice question. I agree with jnelson0612, what he suggested was exactly right and i hope the explanation which he has given was really useful to you.

[spam deleted]

[tim]

Keep your spam off our boards, roo