1. combination of 8 items

Hi I'm trying to figure out how many combinations with the the following 8 items, I can make.
Let's say that I have the following items in two colors each (blank and White): triangle, circle, square and two rectangles with the same color (no white, no black), let's say red.

rules:

1) two items of the same type cannot be together. for instance a black triangle and white triangle cannot be next to each other.

2) any combination starting with a rectangle is not allowed.

3) the first 4 elements of any of the combinations must be identical to the second half (5 to8)but with the opposite color. For instance: black circle, white triangle, black square, rectangle, white circle, black triangle, white square, rectangle.

4) reversed combinations based on color are not needed. for instance if we take the above combination and reverse it:white circle, black triangle, white square, rectangle,black circle, white triangle, black square, rectangle. We only need one of these combinations but not both.

when I'm doing the calculation. I'm thinking of C4,4 = 4x3=12

but because I have to ignore all combinations starting with rectangle. the number should be 8 posible combinations. is this correct?

However, for some reason I think there can't be 8 posible combinations but actually probably only 4.
is this correct?

2. Re: combination of 8 items

how are you going to achieve the opposite color (rule 3) of the rectangle when it only comes in red? Do you just ignore this for the rectangle?

3. Re: combination of 8 items

yes, rule 3 does not apply for the rectangles as it is single color for both rectangle pieces.

4. Re: combination of 8 items

oops. I forgot another rule .

the first 4 elements cannot be the same color.

5. Re: combination of 8 items

given rule (3) rule (1) cannot occur

given that the first 4 elements of a valid combination are of unique shapes, the further specification that their colors will be opposite in the last 4 elements is unnecessary.

Using rules (2) and (3) we have

$6\cdot \dbinom{3}{1} \cdot 4 \cdot 2 = 144$ possible combos that meet these rules.

$6$ comes from 6 choices for selecting the first element.

$\dbinom{3}{1}$ comes from selecting the position of a rectangle.

The rest of the express comes from selecting the remaining elements and should be clear.

Now there will be duplicates according to rule (4) for each of the non-rectangles. This produces $2^3=8$ too many elements and we reduce the count by this factor

to obtain a total count of

$\dfrac {144}{8}=18$ combinations that meet your rules.

$\left( \begin{array}{cccccccc} \{\text{Tri},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Tri},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{W}\} & \{\text{Tri},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} \\ \{\text{Tri},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Tri},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{W}\} & \{\text{Tri},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} \\ \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} & \{\text{Squ},\text{B}\} \\ \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} & \{\text{Cir},\text{B}\} \\ \{\text{Cir},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Cir},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{W}\} & \{\text{Cir},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} \\ \{\text{Cir},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Cir},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{W}\} & \{\text{Cir},\text{B}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} \\ \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{W}\} & \{\text{Squ},\text{W}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} & \{\text{Squ},\text{B}\} \\ \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} & \{\text{Tri},\text{B}\} \\ \{\text{Squ},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Squ},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{W}\} & \{\text{Squ},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} \\ \{\text{Squ},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Squ},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Tri},\text{B}\} & \{\text{Rec},\text{R}\} \\ \{\text{Squ},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{W}\} & \{\text{Squ},\text{B}\} & \{\text{Cir},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} \\ \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{W}\} & \{\text{Cir},\text{W}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Tri},\text{B}\} & \{\text{Cir},\text{B}\} \\ \{\text{Squ},\text{W}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{W}\} & \{\text{Tri},\text{W}\} & \{\text{Squ},\text{B}\} & \{\text{Rec},\text{R}\} & \{\text{Cir},\text{B}\} & \{\text{Tri},\text{B}\} \\ \end{array} \right)$

6. Re: combination of 8 items

Originally Posted by Verdugo
oops. I forgot another rule .

the first 4 elements cannot be the same color.
given that a rectangle must occur in the first 4 elements they cannot all be the same color.

7. Re: combination of 8 items

Originally Posted by romsek
given that a rectangle must occur in the first 4 elements they cannot all be the same color.
sorry what I meant to say is:
the circle, the square and the triangle in the first 4 spots cannot be the same color. There has to be always at least one with a different color.

another rule. the first 4 elements cannot have two figures the same. no two circles, no two square, no two triangles, no two rectangles. in a sense there has to be at least one figure of each.

thx for the help.

8. Re: combination of 8 items

Originally Posted by Verdugo
sorry what I meant to say is:
the circle, the square and the triangle in the first 4 spots cannot be the same color. There has to be always at least one with a different color.

another rule. the first 4 elements cannot have two figures the same. no two circles, no two square, no two triangles, no two rectangles. in a sense there has to be at least one figure of each.

thx for the help.
tell you what

you get all your rules down and repost them all in a new post

I'm not looking at this again until you are happy with your rule set.

9. Re: combination of 8 items

Originally Posted by Verdugo
I'm trying to figure out how many combinations
with the the following 8 items, I can make.

Let's say that I have the following items in two colors each (blank and White): triangle, circle, square and two rectangles with the same color
(no white, no black), let's say red.
Why don't you use something "simpler", like:
black and white letters A, B and C
2 red letter D's

Makes it easier/faster to state CLEAR rules...get my drift?