Sol. 10 IIT students T 1 , T 2 ,........T 10 can be arranged in 10 ! ways. Now the number of ways in which two PET student can be placed will be equal to the number of ways in which 3 consecutive IIT students can be taken i.e. in 8 ways and can be arranged in two ways

⇒ (10 !) (8) (2 !)

Alternatively

3IIT students can be selected in ways. Now each

selection of 3 IIT and two PET students in P 1 T 1 T 2 T 3 P 2 can be arranged in (2 !) (3 !) ways. Call this box X.

Now this X and the remaining & IIT students can be arranged in 8 ! ways

⇒ Total ways = 10 C 3 (2!) (3 !) (8!)]