Question

Seven different teachers are to deliver lectures in seven periods of a class on a particular day. $A$,$B$ and $C$ are three of the teachers. The no.of ways in which a routine for the day can be made such that $A$ delivers his lecture before $B$ and $B$ before $C$ is?

• A. $420$
• B. $120$
• C. $210$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct answer:

Finding the number of ways the routine can be made for seven different teachers such that $A$ delivers his lecture before $B$ and $B$ before $C$:

Since the order of teachers to deliver lectures is $A,B\text{and}C$.

We have to reserve the three slots, then the total number of possible ways is given as: ${}^7{\mathrm{ℂ}}_3$

$$\begin{gathered} {\therefore }^7{\mathrm{ℂ}}_3=\frac{7!}{3!(7-3)!} \\ =\frac{7\times 6\times 5\times 4!}{3!\times 4!} \\ =\frac{7\times 6\times 5}{3\times 2\times 1} \\ =35\text{ways} \end{gathered}$$

Now, the remaining slots are $(7-3)$ or $4$.

The given four slots can be put in $4!$ ways or $24$ ways.

Therefore, the Number of ways routine can be made for seven different teachers such that $A$ delivers his lecture before $B$ and $B$ before $C$, can be given as:

$$\begin{gathered} \Rightarrow 35\times 24 \\ \Rightarrow 840\text{ways} \end{gathered}$$

Therefore, the correct answer is option (D).