Question

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) $ ^{7}{{C}_{2}}\times 6! $
(d) None of these

Answer 1

Hint: Here, we will use the concept of permutation and combination. First of all, we will place the other 6 boys and then the two brothers will be seated in the gaps created. When the 6 boys are seated , a total of 7 gaps are created. The two boys who are brothers can be seated in any of the two gaps from these 7 gaps, in this way they wouldn’t be seated together.We will use the formulas $ ^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ and $ ^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} $ .

Complete step-by-step answer:
Since, we are given that 6 boys are there other than the two boys who are brothers. First of all we will place those 6 boys.
Since, the total number of ways to arrange n different objects is given as n!.
So, the total number of ways in which these 6 boys can be seated is = 6!
Now, when these 6 boys are seated, we have 7 places or gaps where the other two boys can be placed.
If we denote X as the available places for the two boys and B as the acquired places by the 6 boys, then it can be shown as:
 $ X{{B}_{1}}X{{B}_{2}}X{{B}_{3}}X{{B}_{4}}X{{B}_{5}}X{{B}_{6}}X $
So, we can see that the total number of empty spaces are = 7.
Now, we have to choose any two of these empty spaces.
So, the number of ways in which two spaces can be chosen is = $ ^{7}{{C}_{2}} $
The places of these two boys can also be interchanged. So, the total number of ways in which these 2 boys can be seated is = $ ^{7}{{C}_{2}}\times 2! $
Therefore, the total number of ways in which the whole arrangement can be done is:
 $ \begin{align}
  & =6!{{\times }^{7}}{{C}_{2}}\times 2! \\
 & =6!\times \dfrac{7!}{2!\times \left( 7-2 \right)!}\times 2! \\
 & =6!\times \dfrac{7!}{\left( 7-2 \right)!} \\
 & =6!{{\times }^{7}}{{P}_{2}} \\
\end{align} $
Hence, option (b) is the correct answer.

Note: Students should note here that the total number of gaps created after placing n objects is always equal to (n+1) and that the two brothers can be seated in any two of these gaps. A possibility of mistake that could be made by the students is that they do not multiply $ ^{7}{{C}_{2}} $ by 2!. It is necessary because the two brothers can interchange their places. They will not be sitting together even if their places are interchanged. The calculations must be done properly to avoid mistakes.