There are 3 girls in a class of 10 students. | vedclass

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(B) Total students = $10$. Number of girls = $3$. Number of boys = $10 - 3 = 7$.To ensure no two girls are together,we first arrange the $7$ boys in a

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$7! 	imes ^8P_3$

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(B) Total students = $10$. Number of girls = $3$. Number of boys = $10 - 3 = 7$.To ensure no two girls are together,we first arrange the $7$ boys in a row,which can be done in $7!$ ways.This creates $8$ possible gaps (including the ends) where the $3$ girls can be seated: $\_ B_1 \_ B_2 \_ B_3 \_ B_4 \_ B_5 \_ B_6 \_ B_7 \_$.The number of ways to choose and arrange $3$ girls in these $8$ gaps is given by $^8P_3$.Therefore,the total number of arrangements is $7! 	imes ^8P_3$.

There are $3$ girls in a class of $10$ students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is

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