If two different numbers are taken from the set $$\{0, 1, 2, 3, \ldots, 10\}$$; then the probability that their sum as well as absolute difference are both multiple of 4, is:

We have to choose two different numbers from the finite set $$\{0,1,2,3,\ldots ,10\}$$ and we want both their sum and their absolute difference to be multiples of $$4$$.

First, let us count the total number of unordered pairs of distinct numbers that can be chosen from the eleven-element set. The formula for combinations gives

$$^ {11}C_{2} \;=\;\frac{11\times10}{2}=55.$$

So there are $$55$$ possible pairs in all.

Now we must find how many of these pairs satisfy both conditions simultaneously:

$$a+b\equiv0\;(\text{mod }4)\quad\text{and}\quad|a-b|\equiv0\;(\text{mod }4).$$

Because the absolute difference is a multiple of $$4$$, the two numbers must give the same remainder when divided by $$4$$; in other words, they lie in the same congruence class modulo $$4$$. Let us therefore write the common remainder as $$r$$, with

$$r\in\{0,1,2,3\}\quad\text{and}\quad a\equiv r\;(\text{mod }4),\; b\equiv r\;(\text{mod }4).$$

Next, insist that their sum is also a multiple of $$4$$:

$$a+b\equiv r+r=2r\equiv0\;(\text{mod }4).$$

So we need $$2r\equiv0\;(\text{mod }4).$$ Let us examine the four possible values of $$r$$:

$$\begin{aligned} r=0&\;:\;2r=0\equiv0\pmod4\quad\text{(allowed)},\\ r=1&\;:\;2r=2\not\equiv0\pmod4\quad\text{(rejected)},\\ r=2&\;:\;2r=4\equiv0\pmod4\quad\text{(allowed)},\\ r=3&\;:\;2r=6\equiv2\pmod4\quad\text{(rejected)}. \end{aligned}$$

Hence only the two residue classes $$r=0$$ and $$r=2$$ satisfy both conditions. We must therefore list the members of the original set that fall into these two classes:

Residue $$0$$ modulo $$4$$: $$0,4,8$$

Residue $$2$$ modulo $$4$$: $$2,6,10$$

All pairs chosen from within the same class already have a difference divisible by $$4$$, and since the class itself is admissible, their sum is automatically a multiple of $$4$$ as well. Thus every distinct pair drawn from each of these two groups is favourable.

Number of favourable pairs from the residue-$$0$$ group:

$$^3C_2=\frac{3\times2}{2}=3.$$

Number of favourable pairs from the residue-$$2$$ group:

$$^3C_2=3.$$

Total favourable pairs:

$$3+3=6.$$

Finally, the required probability equals the ratio of favourable pairs to total pairs:

$$\text{Probability}=\frac{6}{55}.$$

Hence, the correct answer is Option A.