The number of ways, in which the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}$, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

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Aicharya · Accepted Answer

$\begin{aligned} & =	ext { Total }-\left[\left(\mathrm{All~} \mathrm{in~} R_1 	ext { and } \mathrm{R}_3ight)+\left(	ext { All in } \mathrm{R}_2 	ext { and } \mathrm{R}_3ight)+ \left(	ext { All in } \mathrm{R}_1 	ext { and } \mathrm{R}_2ight)ight] \ & ={ }^8 \mathrm{C}_5 \cdot 5!-\left\{5!+ 5!+{ }^6 \mathrm{C}_5 \cdot 5!ight\} \ & =5!(56-1-1-6)=120(48) \ & =5760\end{aligned}$ .

The number of ways, in which the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}$, D, E can be placed in the 8…

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