Question Number 52381 by Necxx last updated on 07/Jan/19
$${If}\:{all}\:{the}\:{words}\:{with}\:{or}\:{without} \\ $$$${meaning}\:{are}\:{written}\:{using}\:{the} \\ $$$${letters}\:{of}\:{the}\:{word}\:{QUEEN}\:{and} \\ $$$${are}\:{arranged}\:{as}\:{in}\:{English} \\ $$$${dictionary}.{What}\:{is}\:{the}\:{position} \\ $$$${of}\:{the}\:{word}\:{QUEEN}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\mathrm{4}{th}\:{b}\right)\mathrm{45}{th}\:{c}\right)\mathrm{46}{th}\:{d}\right)\mathrm{47}{th} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jan/19
$${total}\:{words}=\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{60} \\ $$$${words}\:{start}\:{Q}=\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{12} \\ $$$${words}\:{start}\:{with}\:{U}=\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{12} \\ $$$${so}\:{total}\:{words}\:{excluding}\:{the}\:{words}\:{start}\:{with} \\ $$$${U}\:{and}\:{Qis}\:=\mathrm{60}−\mathrm{24}=\mathrm{36} \\ $$$${QE}\:{words}=\mathrm{6} \\ $$$$ \\ $$$${QNwords}\:=\frac{\mathrm{3}}{\mathrm{2}!}=\mathrm{3} \\ $$$${QU}\:{words}\rightarrow\mathrm{3}\left({QUEEN},{QUENE},{QUNEE}\right) \\ $$$${so}\:{rank}\:{of}\:{QUEEN}\:{is} \\ $$$$=\mathrm{36}+\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{46}{th} \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 07/Jan/19
$${total}\:{number}\:{of}\:{words}=\frac{\mathrm{5}!}{\mathrm{2}!}=\mathrm{60} \\ $$$${number}\:{of}\:{words}\:{Uxxxx}=\frac{\mathrm{4}!}{\mathrm{2}!}=\mathrm{12} \\ $$$${QUNEE},\:{QUENE}=\mathrm{2} \\ $$$$\mathrm{12}+\mathrm{2}=\mathrm{14} \\ $$$${i}.{e}.\:\mathrm{14}\:{from}\:\mathrm{60}\:{words}\:{are}\:{after}\:{QUEEN}, \\ $$$${i}.{e}.\:{QUEEN}\:{is}\:{the}\:\mathrm{46}{th}\:{word}. \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{c}\right)\:{is}\:{correct} \\ $$