Question

Ten 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 3! \times 8! \\\ (B) \;10! \times 2! \times 3! \times 8! \\\ (C) \;\;{\text{5!}} \times {\text{2!}} \times {\text{9!}} \times {\text{8!}} \\\
$(D)$ None of these.

Answer 1

Hint: In this question the concept of combination will be used like the number of combinations of $N$ different things taken $r$ at a time is ${}^n{C_r}$.

Complete step-by-step solution :
According to a question there are ten IIT and $2$ DCE students sitting in a row that means in row there are $12$ seats .
Hence, according to the condition three IIT students who will be between the IIT students can be selected in ${}^{10}{C_3}$ ways.
Now, two DCE students having three IIT students between them can be arranged in $2!\; \times \;{\text{3!}}$ ways.
Finally, a group of above five students and the remaining seven students together can be arranged in $8!$ ways.
Hence, total number of ways is ${}^{10}{C_3}\; \times \;{\text{2!}}\; \times {\text{3!}}\; \times \;8!$ ways.

Note: In such types of questions first see the conditions that there are $12$ seats in a row and between two DCE exactly three IIT students are sitting so here the concept is the combination i.e. a selection of items from a collection , such that the order of selection does not matter. Hence it is advisable to remember the concept while involving into combination questions.