# How many ways can I color 8 tiles with colors? (easy)

I'm trying to learn the very basics of permutations and combinations and since I'm completely new to this subject I'd like to get a good explanation to the following problems. My task is this:

1) Suppose we have 8 tiles and that we're going to color exactly two of them with blue and exactly two with green. How many way can we do that?

2) Suppose we have eight tiles, but this time we're going to color exactly two tiles blue such that that they're not adjacent. Can a genereral formula be derived from this?

My work so far:

1) The ordered selection of eight tiles is $8!$ and we have to divide this by the number of permutations of colored and non-colored tiles. The number of permutations of the non-colored tiles are $4!$ and $2!$ for each color. If this is correct we should get the following: $$\dfrac{8!}{4!2!2!}=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot1}{4\cdot 3\cdot 2\cdot 1\cdot 2\cdot 1\cdot 2\cdot 1}=7\cdot 6\cdot 5\cdot 2=420$$

I'm not sure how to tackle problem 2). So to sum up, verification of problem 1) and help on problem two, atleast good hints. Also if you have good, relatively short guides (pages/channels/notes etc.) on combinatorics/permutations, feel free to post it. Thanks in advance.

• A more intuitive way (at least for way) of understanding the solution to the first problem is $\binom{8}{2}\binom{6}{2},$ i.e., how many ways can we choose two out of first eight, then six. – Bobson Dugnutt Oct 27 '16 at 7:21

1) If it is so, we subtract from 420 the colorings which have the two blue tiles adjacent. With 8 tiles there are 7 ways to choose 2 adjacent tiles. Then, among the remaining 6 tiles we choose the two tiles to be colored green in $\frac{6!}{4!2!}=15$ ways.
2) If the color green is not used in part 2, we simply subtract from $\frac{8!}{6!2!}=28$ the 7 ways to choose 2 adjacent tiles.