QJEEM09Maedcb: JEE Main 2009 Mathematics - Class 11 JEEM09Maedcb Solution: From 6 different novels and 3 different dictionaries, 4 novels…

From 6 different novels 4 novels can be chosen = $${}^6{C}_4$$ ways

And from 4 different dictionaries 1 can be chosen = $${}^3{C}_1$$ ways

$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C}_4\times {}^3{C}_1$$ ways

Let 4 novels are N₁, N₂, N₃, N₄ and 1 dictionary is D₁.

Dictionary should be in the middle. So the arrangement will be like this

_ _ D₁ _ _

On those 4 blank places 4 novels N₁, N₂, N₃, N₄ can be placed. And 4 novels can be arrange $$4!$$ ways.

$$\therefore$$ Total no of ways = $${}^6{C}_4\times {}^3{C}_1$$$$\times 4!$$ = 1080