Question

In how many ways can 10 math’s teachers and 4 science teachers can be seated at a round table if all the 4 science teachers do not sit together
$
  (a){\text{ 13!}} \times {\text{4!}} \\
  (b){\text{ 13!}} \\
  (c){\text{ 13! - }}\left( {10! \times 4!} \right) \\
  (d){\text{ 10!}} \times {\text{4!}} \\
 $

Answer 1

Hint – In this question we have to make teachers sit in a round table and the condition given is all 4 science teachers do not sit together. Use the concept of round table that if in total n people are to be seated in a round table then the total way of making them sit is$\left( {n - 1} \right)!$. Keep in mind that all 4 science teachers do sit together.

Complete step-by-step answer:
There are 10 math teachers and 4 science teachers.
Total teachers is 14
Total number of ways that they can be seated in a round table is $\left( {n - 1} \right)!$
Where n is the number of teachers.
Therefore total number of ways that they can be seated in a round table is $\left( {14 - 1} \right)! = 13!$
Now consider 4 science teachers as an identity therefore total teachers are $ = 10 + 1 = 11$
So, number of circular arrangement $ = \left( {11 - 1} \right)! = 10!$
Now science teachers can arrange among themselves in $ = \left( {10! \times 4!} \right)$
So, the total number of ways all the science teachers do not sit together is the subtraction of total arrangements and the arrangements of all science teachers sit together.
Therefore possible arrangements are $ = 13! - \left( {10! \times 4!} \right)$
Hence option (c) is correct.

Note – Whenever we face such types of problems the key concept involved is that we have to use the concept of round table arrangements moreover never forgets to make permutations for the arranged teachers. This concept will help you get on the right track to reach the answer.