Problem with poker chips

Question

Dave has $10$ poker chips, $5$ of which are red and the other $5$ 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?

Can someone please help me answer this, I know there is a similar problem out there but it was with $6$ and $4$ instead of $5$ and $5$ so I was unsure how to solve it using odd numbers. Thanks appreciate the help.

Answer 1

There a $10!$ ways to arrange $10$ pieces. But this number includes every single possibility.

Since you have $5$ reds, you can swap a red chip with any other red one and you will get the exact same arrangement. There are $5!$ to arrange the $5$ red chips therefore you divide $10!$ by $5!$.

The same is valid for the whites: divide again $\frac {10!}{5!}$ by $5!$.

Finally you get that the total number of arrangements is $\frac {10!}{5!\times5!}$.

The general formula for the number of different arrangements of $N$ elements, of which $m_1, m_2,...,m_n$ are the same, is $\frac{N!}{m_1!\times m_2!\times...\times m_n!}$

Now, note that every single stack arrangement when flipped has a corresponding arrangement, which is different from it. This is true, save for the symmetric arrangements - they are flipped onto themselves, not a different stack arrangement. Therefore, you are left to calculate how many of the possible stack arrangements are symmetric, divide the rest by $2$, since they are identical when flipped, and you will get the answer. Luckily, there are no symmetric arrangements since you have even number of positions and odd number of chips of the same color.

So you answer would be $\frac {10!}{2\times5!\times5!}$.

Answer 2

So the total way of arranging the chips is 10!/5!5!. Now consider this, if we say let red chips be denoted by 1 and white chips be denoted by 0, then all the palindromes formed will be considered identical as if you flip them you get the same order (eg 1000110001 ). Now all the non palindromic sequences are considered twice in the 10!/5!5! (as 10!/5!5! counts 110010001,100010011, but as a rule these are identical) Thus we need to divide the whole thing by 2, But in doing so we divide number of palindromic sequences which aren't counted twice by 2, hence we first add the number of palindromic sequence to the total way and divide the whole by 2 to get the answer the the should be : {[10!/5!5!]+no of palindromic sequence}/2. But here we understand the the palindrome should be using 5 ones and 5 zeros which isn't possible since 10 is even and you need even numbers of ones and zeros adding to 10 (like 6 and 4 as suggested), thus no of palindromic sequence = 0 thus the final ans : {10!/5!5!}/2.