AIEEE 2009 Solved Paper

From 6 different novels 4 novels can be chosen $= {}^6C_4$ waysAnd from 4 different dictionaries 1 can be chosen $= {}^3C_1$ ways$\therefore$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen $= {}^6C_4 \times {}^3C_1$ waysLet 4 novels are $N_1, N_2, N_3, N_4$ and 1 dictionary is $D_1$.Dictionary should be in the middle. So the arrangement will be like this$--D_{1--}$On those 4 blank places 4 novels $N_1, N_2, N_3, N_4$ can be placed. And 4 novels can be arrange 4 ! ways.$\therefore$ Total no of ways $= {}^6C_4 \times {}^3C_1 \times 4! = 1080$