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Q. Seven different lectures are to be delivered in 7 period of a class on a particular day. Out of 7 lecturers $A , B$ and $C$ are three of them. If the number of ways in which a routine for the day can be made such that 'A' deliver his lecture before ' $B$ ' and ' $B$ ' before 'C 'is $N$, then find the value of $\left(\frac{ N }{120}\right)$.

Permutations and Combinations

Solution:

Select 3 places for $A , B , C$ in ${ }^7 C _3$ way $\times \times \times \times \times \times$
and arrange them only in 1 way.
Remaining 4 places can be arranged in 4 ! ways
$\therefore$ Total ways $={ }^7 C _3 \cdot 4 !=35 \cdot 24=840$. $\Rightarrow N =840$
$\Rightarrow N =840$
$\therefore\left(\frac{ N }{120}\right)=\frac{840}{120}=7$