To solve this question, we need to use the formula for combinations, which is given by:

nCr = n! / (r! * (n-r)!)

where n is the total number of items, and r is the number of items being chosen.

Given that 20Cn2 = nC16, we can set up the equation as follows:

20! / (n! * (20-n)!) = n! / (16! * (n-16)!)

Cancelling out the common terms n! and (20-n)! on both sides of the equation, we get:

20! / (20-n)! = 1 / (16! * (n-16)!)

Now, we can simplify the equation further:

(20-n)! = (16! * (n-16)!)

Taking the factorial of both sides, we have:

(20-n) * (20-n-1) * (20-n-2) * ... * 1 = (n-16) * (n-16-1) * (n-16-2) * ... * 1

Expanding both sides, we get:

(20-n) * (19-n) * (18-n) * ... * 1 = (n-16) * (n-17) * (n-18) * ... * 1

Now, we can cancel out the common terms on both sides:

(20-n) * (19-n) * (18-n) = (n-16) * (n-17) * (n-18)

Expanding both sides, we get:

(20n^2 - 380n + 3420) = (n^3 - 51n^2 + 816n - 4896)

Simplifying the equation, we have:

n^3 - 51n^2 + 816n - 4896 - 20n^2 + 380n - 3420 = 0

n^3 - 71n^2 + 1196n - 8316 = 0

Now, we can find the roots of this equation using the Rational Root Theorem or any other suitable method. After solving the equation, we find that it has no rational roots. Therefore, the correct answer is option 'D' (none of these).

Note: It is important to remember that this solution assumes that n is a positive integer. If n can be any real number, then the answer would be different.