@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! ?
