The math is rather simplistic on this, as we start with something as simple as knowing the row we are in. As the recursive definition states, A0 = .5 (signifying the zero row, or the top row of Pascal’s Triangle), we are left with a rather simple recursion in the note of An = An * 1/2, where n signifies the row, and for every integer, you have a logical number of pairs, and every fractional half, you have the whole number of pairs and the simple “pair” upon itself. However, there’s another feature to finding the amount of pairs within each row of Pascal’s Triangle.
Since Pascal’s Triangle uses combinations to find each value of the binomial theorem, we discover by adding the placement of the column, i.e. 0, 1, 2, 3, 4, 5 for the 6th row, we discover by using the formula n(n-1), divided by the summation of the columns, we discover 2, as in the basis for the famous triangle and the theorems therein.
Using our simple example (the 6th row), we discover n(n-1) = 6*5, or 30, and then divide it by the summation of the column numbers (0+1+2+3+4+5 = 15), we discover the number 2. The proof goes as follows:
N(N-1)/(N(N-1)/2) => 2(N(N-1))/(N(N-1)) => 2/1 => 2. First step, we multiply by the reciprocal of ½, so as to allow for a whole denominator. We are then left with 2 times the quantity of N times N minus 1, divided by the quantity of N times N minus 1. Since the quantity N times N minus 1 is divided by a quantity of the exact same measure, we are left with 2 divided by 1, or the number 2.
Or, rather, 0C5, 1C5, 2C5, 3C5, 4C5, 5C5 translates to 5+5+5+5+5+5 over 0+1+2+3+4+5, for the value of 2.