Question

If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is
(a) 40320
(b) 2520
(c) 20160
(d) none of these

Answer 1

Hint: Here, we will find the two possible cases where the first speaker will address before another speaker. We will assume two speakers as ${{s}_{1}}$ , ${{s}_{2}}$. So, we will get two possibility as if ${{s}_{1}}$ speaks before ${{s}_{2}}$ which is written as ${{s}_{1}} > {{s}_{2}}$ and remaining 6 speakers can speak anywhere in meeting. The second case is: if ${{s}_{2}}$ speaks before ${{s}_{1}}$ which is written as ${{s}_{2}} > {{s}_{1}}$. Then, we will find a total number of ways without any condition that will be $8!$. Then, we divide the total possibility by $2!$ and we will get the answer.

Complete step-by-step answer:
Here, we are told that there are a total of 8 speakers who will address the meeting out of which there will be 2 speakers who will speak before another speaker speaks.
So, let us assume any two speakers as ${{s}_{1}}$ , ${{s}_{2}}$. We have now two cases i.e. first: if ${{s}_{1}}$ speaks before ${{s}_{2}}$ which is written as ${{s}_{1}} > {{s}_{2}}$ and remaining 6 speakers can speak anywhere in meeting. The second case is: if ${{s}_{2}}$ speaks before ${{s}_{1}}$ which is written as ${{s}_{2}} > {{s}_{1}}$ .
Now, we will find total ways in which 8 speakers will speak in a meeting without any conditions. So, this will be equal to $8!$.
On solving this, we will get value as
$8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$
Now, we have two possibilities as we saw above. So, we will divide the answer by $2!$. We will get as
$\dfrac{8!}{2!}=\dfrac{40320}{2\times 1}=20160$
Thus, the answer where both the condition is satisfied is 20160 ways where either ${{s}_{1}} > {{s}_{2}}$ or ${{s}_{2}} > {{s}_{1}}$.

So, the correct answer is “Option c”.

Note: Another approach to solve the problem is by taking two speakers ${{s}_{1}}$, ${{s}_{2}}$. Now, let us say that ${{s}_{1}}$ speaks in first position then ${{s}_{2}}$ have to choose from second to eighth position which will be 7 ways. So, the remaining 6 speakers have 6 ways to address each other in a meeting. It can be written as $=7\times 6!$.
Now, if we say that ${{s}_{1}}$ speaks in second position, then ${{s}_{2}}$ will have a third to eight position to speak i.e. 6ways to speak. Remaining speakers will have to speak in 6 ways. It can be written as $=6\times 6!$.