Recently I was trying to count the number of n-bit numbers that have an equal number of 1’s and 0’s. For example, there are 6 such numbers for 4-bit long numbers (0011, 0101, 0110, 1001, 1010, 1100). I wrote a simple program to count such numbers for arbitrarily long numbers. However, the program was computationally complex and it took a long time to compute the result for large values of n. After spending some time trying to figure out the equation myself, I searched the internet and stumbled upon Catalan numbers. I determined that the equation is related to Catalan numbers. I included the equation below because I couldn’t find it anywhere else. Also, I don’t have a formal proof, but it seems to work. Enjoy!
Number of n-bit numbers that have an equal number of 1’s and 0’s = (m+1) * Cm, where m = n/2 and Cm = mth Catalan number