Question

5 girls including Preeti and Shreya, 5 boys including Mohit and Ninaad are to be seated in a row of 10 seats such that both Preeti and Shreya are seated immediately next to Mohit and Ninaad is not immediately next to either Preeti or Shreya. The total number of ways to arrange all the 10 boys and girls is

• A. $12\times 7!$
  
• B. $64\times 6!$
  
• C. $240\times 6!$
  
• D. $80\times 6!$

Answer 1

The correct option is A $12\times 7!$



We shall denote given names using first letter only. If both P and S are immediately next to M, then M cannot be at positions 1 or 10. So M can be seated at 8 locations {2, 3,….9}. After fixing M, P and S can be permuted on his either side in 2! Ways. The unit ‘PMS’ has 8 possible positions {123, 234, …… 8910}. Now in order to separate N, we need to consider 2 cases.
If PMS is at ‘123’ or ‘8910’ then N can take 6 locations (note that only one seat needs to be ignored). Hence when the unit PMS is at extreme ends,
Possible no. of ways $\left(N_1\right)=\underset{\underset{PMSPosition}{\uparrow }}{2}\times \underset{\underset{P\&SPermutaion}{\uparrow }}{2}\times \underset{\underset{N^{\prime }sPosition}{\uparrow }}{6}\times \underset{\underset{RemainingPermutation}{\uparrow }}{6!}$
$N_1=24\times 6!$
There are 6 remaining locations for ‘PMS’ viz {234, 345,……789}. In this case N has only 5 options as 2 seats need to be ignored.
Possible no. of ways $=\left(N_2\right)=\underset{\underset{PMSPosition}{\uparrow }}{6}\times \underset{\underset{P\&SPermutaion}{\uparrow }}{2}\times \underset{\underset{N^{\prime }sPosition}{\uparrow }}{5}\times \underset{\underset{RemainingPermutation}{\uparrow }}{6!}$
$N_2=60\times 6!$
Hence, total required ways = $N_1+N_2=84\times 6!=12\times 7!$