Number of ways to stack poker chips if two stacks that can be obtained from each other by reflection are considered to be equivalent? Dave has 10 poker chips, 6 of which are red and the other 4 of which are white. Dave likes to stack his chips and flip them over as he plays. How many different 10-chip stacks can Dave make if two stacks are not consider distinct if one can be flipped to appear identical to the other?
So I know that the number of ways to arrange the poker chips is 10 choose 6. But what do I do after that to satisfy the second condition?
 A: You subtract the number of "symmetric" stacks (ie. flipped = itself) to get the number of asymmetric ones. But since they get double counted, divide by 2.
Number of symmetric stacks = $\binom 5 3$ since each half would have 3 red and 2 white.
$$\frac12 (\binom{10}6 -\binom53)+\binom53=110$$
A: Let's consider a simpler problem first.  

In how many ways can four red and two white poker chips be stacked if two stacks are considered equivalent if they can be obtained from each other by reflection?  

There are $\binom{4 + 2}{2} = \binom{6}{2} = 15$ ways to arrange four red and two white poker chips in a row since the sequence is completely determined by which two of the six positions are occupied by the white chips.  If we simply divide this number by $2$, we do not get an integer.  The reason for this is that $3$ of the arrangements are symmetric with respect to the center of the sequence.  In the list below, we match each sequence with its reflection.
$$RRRRWW \longleftrightarrow WWRRRR$$
$$RRRWRW \longleftrightarrow WRWRRR$$
$$RRRWWR \longleftrightarrow RWWRRR$$
$$RRWRRW \longleftrightarrow WRRWRR$$
$$RRWRWR \longleftrightarrow RWRWRR$$
$$\color{blue}{RRWWRR \longleftrightarrow RRWWRR}$$ 
$$RWRRRW \longleftrightarrow WRRRWR$$
$$\color{blue}{RWRRWR \longleftrightarrow RWRRWR}$$
$$\color{blue}{WRRRRW \longleftrightarrow WRRRRW}$$
The three sequences highlighted in blue are their own reflections.  Notice that there are a total of nine rows.  This is because there are six pairs of distinct sequences that can be obtained from each other by reflection and three sequences that are their own reflections.  
How could we calculate this without listing all the sequences?
Why are there three sequences that are symmetric with respect to the center of the sequence?  
Each such sequence has two red and one white chip on each side of the center of the sequence.  Choosing the position of the white chip that is to the left of the center determines the position of the white chip to the right of the center.  Therefore, there are $\binom{2 + 1}{1} = \binom{3}{1} = 3$ such sequences.   
The remaining $15 - 3 = 12$ sequences that are not their own reflections form $6$ pairs of sequences that can be obtained from each other by reflection.  
Hence, the number of stacks we can obtain using four red and two white poker chips if two stacks that can be obtained from each other by reflection are considered to be equivalent is 
$$\frac{1}{2}\left[\binom{6}{2} - \binom{3}{1}\right] + \binom{3}{1} = \frac{1}{2}(15 - 3) + 3 = \frac{1}{2} \cdot 12 + 3 = 6 + 3 = 9$$
where the first term counts the number of pairs of distinct sequences that can be obtained from each other by reflection and the second term counts the number of sequences that are their own reflections.  

In how many ways can six red and four white poker chips be stacked if two stacks are considered equivalent if they can be obtained from each other by reflection?

There are $\binom{6 + 4}{4} = \binom{10}{4}$ sequences that can be formed using six red and four white poker chips.
A sequence that is symmetric with respect to its center will have three red and two white poker chips on each side of its center.  Choosing the positions of the white chips to the left of the center determines the positions of the white chips to the right of the center.  Hence, there are $\binom{3 + 2}{2} = \binom{5}{2}$ such sequences.
Therefore, there are $\binom{10}{4} - \binom{5}{2}$ sequences that are not their own reflections.  Hence, there are $\frac{1}{2}\left[\binom{10}{4} - \binom{5}{2}\right]$ pairs of distinct sequences that can be formed by matching a sequence with its reflection.  
To obtain the answer, we must add the number of pairs of distinct sequences that can be obtained from each other by reflection to the number of sequences that are their own reflections.  Hence, the number of stacks of six red and four white poker chips that can be obtained if two stacks are considered equivalent if they can be obtained from each other by reflection is 

 $$\frac{1}{2}\left[\binom{10}{4} - \binom{5}{2}\right] + \binom{5}{2}$$

