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  • Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:a)Boys and girls alternate i)5!x6!

Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
\begin{array}{|l|l|l|l|} \hline \mathbf{C}_{\mathbf{1}} & & \mathbf{C}_{2} & \\ \hline \text { (a) } & \begin{array}{l} \text { Boys and girls } \\ \text { alternate: } \end{array} & \text { (i) } & \begin{array}{l} \text { 5! } \times \\ \text { 6! } \end{array} \\ \hline \text { (b) } & \begin{array}{l} \text { No two girls sit } \\ \text { together : } \end{array} & \text { (ii) } & \begin{array}{l} 10 ! \\ -5 \\ !6 ! \end{array} \\ \hline \text { (c) } & \begin{array}{l} \text { All the girls sit } \\ \text { together } \end{array} & \text { (iii) } & \begin{array}{l} (5 !)^{2} \\ + \\ (5 !)^{2} \end{array} \\ \hline \text { (d) } & \begin{array}{l} \text { All the girls are } \\ \text { never together : } \end{array} & \text { (iv) } & \begin{array}{l} 2 ! \\ 5 ! \\ 5 ! \end{array} \\ \hline \end{array}
 

Answers (1)

 Given that number of boys=5 and number of girls=5   

a.Boys and girls alternate=\left (5!. 5! \right )+\left (5!. 5! \right )=\left (5! \right )^2+\left (5! \right )^2

b.No two girls sit together=5!. 6!

c.All the girls sit together=2! . 5!. 5!

d .All the girls are never together :  

 Total number of boys and girls=5+5=10

 Number of ways=10!-5!6!

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