JEE MAIN - Mathematics (2018 (Offline) - No. 1)
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always in the middle. The number of such
arrangements is :
at least 750 but less than 1000
at least 1000
less than 500
at least 500 but less than 750
Explanation
From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways
And from 3 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
And from 3 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
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