The value of r for which 20C20rC0 + 20C20rC1 + 20C20rC2 + .... 20C20rCr is maximum, is

Let the given expression be denoted by S. Then
S = ∑{k=0}^{r} ‖C(20, k)C(20, r-k)
Using the formula for the sum of products of combinations:
∑{k=0}^{r} C(m, k)C(n, r-k) = C(m+n, r)
In our case, m = 20, n = 20, so we have
S = C(20+20, r) = C(40, r)
The maximum value of C(n, r) occurs when r = n/2 if n is even and r = (n-1)/2 or (n+1)/2 if n is odd.
Since n = 40 is even, the maximum value of C(40, r) occurs at r = 40/2 = 20.
Therefore, the value of r for which the given expression is maximum is 20.