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More # 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$(a){\text{ 13!}} \times {\text{4!}} \\ (b){\text{ 13!}} \\ (c){\text{ 13! - }}\left( {10! \times 4!} \right) \\ (d){\text{ 10!}} \times {\text{4!}} \\$

Last updated date: 22nd Mar 2023
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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.

Total number of ways that they can be seated in a round table is $\left( {n - 1} \right)!$
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 $= \left( {10! \times 4!} \right)$
Therefore possible arrangements are $= 13! - \left( {10! \times 4!} \right)$