$10\text{ IIT \& }2\text{ PET students sit in a row. The total number of ways in which exactly }3\text{ IIT students sit }$

$\text{ between }2\text{ PET students is }K\text{ then }\frac{K}{1000}\text{ is }$

$10\text{ IIT students }{T}_1,T_2,\dots ..T_{10}\text{ can be arranged in }10!\text{ ways. Now the number of ways in which }$

$\text{ two PET student can be placed will be equal to the number of ways in which }3\text{ consecutive IIT }$

$\text{ students can be taken i.e. in }8\text{ ways and can be arranged in two ways }\mathrm{D}(10!)\left(8\right)(2!)\text{. }$

$\text{ Alternatively }3{\text{ IIT students can be selected in }}^{10}{\mathrm{C}}_3\text{ ways. Now each selection of }3\mathrm{IIT}\text{ and }2\mathrm{PET}$

$\text{ students in }{\mathrm{P}}_1{\mathrm{T}}_1{{\mathrm{T}}_2\mathrm{T}}_3{\mathrm{P}}_2\text{ can be arranged in (2!) (3 !) ways. Call this box }\mathrm{X}\text{. Now this }\mathrm{X}\text{ and }$

$\begin{array}{l}\text{ the remaining IIT students can be arranged in }8!\text{ ways }\\ \left.\Rightarrow \text{ Total ways }10{\mathrm{C}}_3(2!)(3!)(8!)\right]\end{array}$