The value of $$r$$ for which $${}^{20}C_r \cdot {}^{20}C_0 + {}^{20}C_{r-1} \cdot {}^{20}C_1 + {}^{20}C_{r-2} \cdot {}^{20}C_2 + \ldots + {}^{20}C_0 \cdot {}^{20}C_r$$ is maximum, is:

We begin with the expression

$$S(r)= {}^{20}C_r \cdot {}^{20}C_0 + {}^{20}C_{r-1} \cdot {}^{20}C_1 + {}^{20}C_{r-2} \cdot {}^{20}C_2 + \ldots + {}^{20}C_0 \cdot {}^{20}C_r.$$

The general term of this finite sum can be written as $$^{20}C_{r-k}\,^{20}C_k$$ where the index $$k$$ goes from $$0$$ up to $$r$$. Hence we may rewrite the whole sum compactly as

$$S(r)=\sum_{k=0}^{r} {}^{20}C_{r-k}\,{}^{20}C_k.$$

Now we recall (and explicitly state) a standard identity that comes directly from the Binomial Theorem:

$$\bigl(1+x\bigr)^{m}\bigl(1+x\bigr)^{n}=(1+x)^{\,m+n}.$$

When we expand each factor separately and then multiply, the coefficient of $$x^t$$ on the left-hand side is obtained by convolution of the two individual coefficient sequences. In symbols,

$$\bigl[(1+x)^m\bigr]\bigl[(1+x)^n\bigr]=\sum_{t=0}^{m+n}\Bigl[\sum_{k=0}^{t}{}^{m}C_k\,{}^{n}C_{t-k}\Bigr]x^{\,t}.$$

Comparing this with the right-hand side $$ (1+x)^{m+n} $$ we see that

$$\sum_{k=0}^{t}{}^{m}C_k\,{}^{n}C_{t-k}={} ^{\,m+n}C_t. \quad -(★)$$

This is the convolution identity for binomial coefficients.

In our problem we have $$m=20,\; n=20$$ and $$t=r$$. Therefore, by direct substitution into identity (★) we get

$$S(r)=\sum_{k=0}^{r}{}^{20}C_k\,{}^{20}C_{r-k}={} ^{\,40}C_r.$$

So the given sum is precisely the single binomial coefficient $$^{40}C_r$$.

Next we determine for which integer $$r$$ (restricted here to $$0\le r\le 20$$ because $$^{20}C_{r-k}$$ vanishes whenever the upper index $$20$$ is smaller than the lower one) the value of $$^{40}C_r$$ is largest.

We use another well-known fact about binomial coefficients:

For a fixed positive integer $$N$$, the sequence $$^{N}C_0,\,^{N}C_1,\,^{N}C_2,\ldots,\,^{N}C_N$$ increases strictly up to $$r=\bigl\lfloor N/2\bigr\rfloor$$ and then decreases symmetrically, with

$$^{N}C_r=^{N}C_{N-r}.$$

Here $$N=40$$ is even, so the unique largest term is achieved at exactly

$$r=\dfrac{N}{2}=20.$$

Since all permissible $$r$$ lie between $$0$$ and $$20$$ (inclusive), the maximum clearly occurs at this central value itself.

Hence, the correct answer is Option B.