In a group of boys, two boys are brothers and six more boys are present in the group. In how many ways can they sit if the brothers are not to sit along with each other?
A. $2\times 6!$
B. ${}^{7}{{P}_{2}}\times 6!$
C. \[{}^{4}{{C}_{2}}\times 6!\]
D. None of these

Hint: We will first start by drawing a rough situation depicting how the boys can be arranged. Then we will try to fit the two brothers in between the space of 6 boys. So, that the brothers are separated and finally we will permute them to find the total ways.

Complete step-by-step answer:
Now, we have a group of boys in which 2 are brothers and size more boys are there. So, if we let the size boys as ${{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}},{{B}_{5}},{{B}_{6}}$ then we have the seating arrangement as,
$X\ {{B}_{1}}\ X\ {{B}_{2}}\ X\ {{B}_{3}}\ X\ {{B}_{4}}\ X\ {{B}_{5}}\ X\ {{B}_{6}}\ X$
Where X denotes a seat in between 6 boys.
Now, since we have to make two brothers sit separately. So, we can choose any 2 of the 7 empty seats and permute them.
Now, we know the ways of permuting 2 objects among 7 is ${}^{7}{{P}_{2}}$.
Also, the ways in which rest 6 boys can sit is 6!.
Therefore, the total way of seating is ${}^{7}{{P}_{2}}\times 6!$.
Hence, the correct option is (B).

Note: To solve this question we have used a trick of first sitting all the persons in which there is no condition and then making the persons which cannot sit together in between. So, they cannot sit together.