Question

A round table conference is to be held among 20 delegates of 20 countries. The number of ways they can be seated if two particular delegates never sit together.
1. \[17\times 18!\]
2. \[18\times 19!\]
3. \[\dfrac{20!}{2}\]
4. \[19!\times 2\]

Answer 1

Hint: A round table conference with 20 participants from 20 different countries is organized. If two delegates are never seated together, the number of different ways they can be seated. As a result, 2 of the 20 delegates must be removed, bringing the total number of delegates to 18. Find the total number of arrangements for the 18 members, excluding two specific members.

Complete step-by-step solution:
We are given that a round table conference is to be held on 20 delegates from 20 countries. We are finding many ways in which they cannot sit together.
Total no. of delegates \[=20\]
2 delegates cannot sit together, that means arrangements of 18 members except two particular delegates.
Now, total no. of delegates \[=18\]
18 person can be seated in \[(n-1)!\]ways where \[n=18\]
\[(n-1)!\]
\[=(18-1)!\]
\[=17!\]
Number of arrangements of two particular members in between 18 places \[{{=}^{18}}{{P}_{2}}\]ways.
\[=18\times 17\]
\[\therefore \] Total arrangements\[=17!\times 18\times 17=17\times 18!\] [where $n\times n-1!=n!$]
Therefore, from among the options given in the question option A is correct which is \[17\times 18!\]
So, the correct option is “option A”.

Note: A permutation is a method of arranging objects in a specific order. Combinations are the methods by which they choose objects from a collection or a group of objects. Combination should be used when the order of the objects isn't important, and Permutation should be used when the order is important. Permutation and combination should not be confused.