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?

A.24
B.32
C.48
D.60
E.192

Thank you Anurag, I understand that I have to multiply 8*6*4, but to divide by 3! is to delete the overcounting numbers, right? Could you be more specific about 3! ?

@mysseo, in your selection of 1st, 2nd and 3rd cars with two different models among 4 colors you may select the same models of cars, as correctly notes Anurag, with the same color but in different order. That is you may select 6 sets of exactly the same type of cars (A or B) with the same color, but do it variously (in different order). The result will be the same always as you end selecting the same types of cars with the same colors. And you must say that this selection is overestimated by 6, because you used so called permuted order (arranged order) for 8, 6 and 4 possible choices

>>> likewise with a pair of six-sided dice: simple counting is 6x6=36 dice side sets, permutation is 6*5=30 dice side sets (excluding the identical sides, 6 pairs 1:1,2:2,3:3, ... 6:6) and combination 6*5/2!=15 because in arranged/permuted order you have 6P2=6!/4! and with the combination (not permuted/arranged order) you get 6C2=6!/(4!*2!) or the number of ways to select different dice side sets excluding 6 pairs (in total there would be 15+6=21 different dice side sets with the pairs included). The difference between ordered and not ordered number of sets (ways) would be 2! (k element number), e.g. 1:3 and 3:1 two sets with the same dice sides <<<

We would be interested in the combination, therefore we divide everything by k (number of elements to be selected), 3! as k=3. This way you turn the permutation into combination, 4C3= 4P3/3!

[spoiler]counting->permutation->combination turnover[/spoiler]

mysseo wrote:Thank you Anurag, I understand that I have to multiply 8*6*4, but to divide by 3! is to delete the overcounting numbers, right? Could you be more specific about 3! ?

Can you please explain why it is not 8*6*4/5!3!?

It might be helpful to think of a simpler scenario. Imagine, for example, that you have 5 different cars to choose from: Red, Blue, Green, Yellow, or White. You want to pick 3 different colored cars, and you want to know how many combinations of cars you can select. If you plug into the combination formula, you'll get 5!/(3!*2!). Answer comes to 10.

However, another way to think about it is to imagine you have 3 slots to fill. You can pick any of the cars for the first slot and so have 5 options. You have 4 options for the second slot, and then 3 options remaining for the third. So far we have 5*4*3. But we're not finished because selecting the Blue, Red, and Green cars is the same as selecting the Red, Green, and Blue cars. Order doesn't matter. But if order did matter, there would be 3! ways to arrange these three elements, so we then have to divide by 3! to make sure we're not counting duplicate scenarios. Put another way, we need to divide by (# interchangeable slots!) Here, there are three interchangeable slots. The answer ends up being (5*4*3)/(3!) Again, we get 10.

In the case of the problem you're asking about, we're not using a conventional formula. Our reasoning is more in line with the second approach outlined above. We had 8 options for the first slot, 6 for the second and 4 for the third. Because there are three interchangeable entities, we divide by 3! I think your confusion comes from trying to apply the conventional formula where it isn't appropriate.

gogmat2015 wrote:Can you please explain why it is not 8*6*4/5!3!?

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https://www.beatthegmat.com/pls-clear-my ... 85743.html