Question

The number of ways in which 10 boys can take positions about a round table if two particular boys must not be seated side by side is:
\[
  A.{\text{ 7}}\left( {8!} \right) \\
  B.{\text{ 9}}\left( {8!} \right) \\
  C.{\text{ 7}}\left( {9!} \right) \\
  D.{\text{ None of the above}} \\
 \]

Answer 1

Hint: In order to solve such a type of problem use the concept of permutation and combination along with formula for arrangements for round tables. First arrange the boys apart from those two boys and then move on to arranging the boys who cannot sit together.

Complete step-by-step answer:

In the problem we have 10 boys to be arranged on a round table out of which 2 of them cannot sit together.
First let us leave those 2 boys and arrange 10 – 2 = 8 boys.
As we know that in round table arrangement of people. If we have to arrange n differently then the total number of arrangements will be $\left( {n - 1} \right)!$.
Number of arrangements of 8 boys on round table $ = \left( {8 - 1} \right)! = 7!$ .
Now as we have arranged 8 boys on the table there are 8 gaps present in the table. 1 gap between each consecutive boy. Now these 2 boys can be seated on any of the 2 seats.
As we know that the number of arrangements of m different items from a group of n items can be done in ${}^n{P_m}$ ways.
So arrangement of 2 boys out of 8 seats can be done in ${}^8{P_2}$ ways.
As these two arrangements are part of one event so the number of arrangements of boys can be found out by the product of these two results.
So number of arrangements
$
   = 7! \times {}^8{P_2} \\
   = 7! \times \dfrac{{8!}}{{\left( {8 - 2} \right)!}}{\text{ }}\left[ {\because {}^m{P_n} = \dfrac{{m!}}{{\left( {m - n} \right)!}}} \right] \\
   = 7! \times \dfrac{{8!}}{{6!}} \\
$
Now let us solve the result in order to get the final result by the use of formulas for the factorial.
$
   = 7! \times \dfrac{{8 \times 7 \times 6!}}{{6!}} \\
   = 7! \times 8 \times 7 = \left( {7! \times 8} \right) \times 7 \\
   = \left( {8!} \right) \times 7 = 7\left( {8!} \right) \\
 $
Hence, number of arrangements of 10 boys on round table out of which 2 are not sitting together is $7\left( {8!} \right)$
So, option A is the correct option.

Note: This problem can also be solved in another way first by finding the total number of arrangements for 10 boys removing the condition and then subtracting from it the number of ways of arrangements in which two boys are always made to sit together by considering them as one person. Students must remember the formula for arrangements of objects in round position, which is also mentioned in the solution.