**Concept: The question tests on applications in Combinatorics**

**Solution:**

Total number of committees that can be formed:

Choose 4 for the first committee out of 12, then choose another 4 out of the remaining 8 and then the last 4 will go into the last committee = 12C4 * 8C4 * 4C4

= 495 * 70 * 1

= 34650

Let Bill and Mitt be already there in the 1st committee. Then 2 more can be chosen out of 10 in 10C2 ways.

Then choose 4 for the next committee out of 8 in 8C4 ways and the last 4 go into the last committee in 4C4 ways.

= 10C2 * 8C4 * 4C4

= 3150

Similarly Bill and mitt can be chosen in the 2nd sub-committee in 3150 ways (10C4 * 6C2 * 4C4) and in another 3150 ways when chosen into the last committee (10C4 * 6C4 * 4C4).

Total combinations where Bill and Mitt are always together = 3150 * 3 = 9150

Total Number of ways in which Bill and Mitt are never together = 34650 - 9150 = 2500

% = 25200/34650∗100

=800/11

=72 8/11** Option D**