Find the number of ways in which two American | vedclass

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(C) Total number of people = $2$ (Americans) + $2$ (British) + $1$ (Chinese) + $1$ (Dutch) + $1$ (Egyptian) = $7$ people.Let the people be $A_1, A_2$

Vedclass · Accepted Answer

$336$

Vedclass · Answer

(C) Total number of people = $2$ (Americans) + $2$ (British) + $1$ (Chinese) + $1$ (Dutch) + $1$ (Egyptian) = $7$ people.Let the people be $A_1, A_2$ (Americans),$B_1, B_2$ (British),$C$ (Chinese),$D$ (Dutch),and $E$ (Egyptian).We need to arrange them in a circle such that $A_1, A_2$ are not together and $B_1, B_2$ are not together.First,arrange the $5$ people $(C, D, E, A_1, B_1)$ in a circle in $(5-1)! = 4! = 24$ ways.There are $5$ gaps created between these $5$ people.We need to place $A_2$ and $B_2$ in these $5$ gaps such that they are not adjacent to their respective nationals.Using the principle of inclusion-exclusion or gap method,the total arrangements where $A_1, A_2$ are separated and $B_1, B_2$ are separated is $336$.

Find the number of ways in which two Americans,two British,one Chinese,one Dutch,and one Egyptian can sit at a round table so that persons of the same nationality are separated.

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