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Given that, the numbers $1,2,3,4,5,6,7,\text{and}\; 8.$

$3,5$ should be present in every group. And $7,8$ never present together in any of the groups.

There are $3$ possibilities.

- Only $7$ is selected, not $8$ is selected $3,5,7$ are already selected, $8$ is not included. We have a choice for $1,2,4, \text{and} 6.$ Each of these numbers is selected or neither be selected. So, the number of ways $= 2 \times 2 \times 2 \times 2 = 2^{4} = 16$
- Only $8$ is selected, not $7$ is selected. Similarly, the number of ways these numbers can be selected or neither be selected $= 2^{4} = 16$
- $3.$ Neither $7$ nor $8$ is selected. $3,5$ are already selected, and $7,8$ are not be selected. Now, we have choices for $1,2,4, \text{and} \;6.$ These numbers can be selected or neither be selected. The number of ways $2^{4} = 16.$ But the case when neither any number be selected, this can’t form three or more digit distinct number. So, the number of ways $= 16-1 = 15$

$\therefore$ The number of groups of three r more numbers $= 16+16+15 = 47.$

Correct Answer $: 47$

$\textbf{Short method:}$

Given that, set of numbers $\{1,2,3,4,5,6,7,8\}$

From the above set, $\{3,5\}$ is taken, then the number of possible subsets containing at least three numbers $= 2^{6}-1 = 64-1 = 63$

From these $63,$ we need t remove the subsets which have $\{3,5,7,8\}.$

The remaining numbers are $\{1,2,4,6\}.$

For these numbers, we have two possibilities, either selected or not selected.

So, the number of subsets possible $= 2^{4} = 16$

$\therefore$ The total number of ways $= 63-16=47.$