Question

$10$ IIT and $2$ DCE students sit in a row. The number of ways in which exactly $3$ IIT students sit between $2$ DCE students is

• A. ${}^{10}{C}_3\times 2!\times 8!$
• B. $10!\times 2!\times 3!\times 8!$
• C. $5!\times 2!\times 9!\times 8!$
• D. $10!\times 2!\times 8$

Answer 1

Answer: (D)

$10!\times 2!\times 8$

There should be $3$ students that has to be placed between $2$ DCE students.
The DCE students can be arranged in $8$ ways, i.e. at position $(1,5),(2,6),(3,7),...(8,12)$
and can interchange their places in $2!$ ways.
Now the rest IIT studentsa can be arranged in${}^{10}{P}_{10}$ ways
$\therefore$ Total ways are $8\times 2!\times {}^{10}{P}_{10}=8\times 2!\times 10!$