Question Number 6090 by sanusihammed last updated on 13/Jun/16
Answered by Rasheed Soomro last updated on 13/Jun/16
$$\:^{{x}+\frac{{y}}{\mathrm{2}}} \mathrm{C}_{\mathrm{2}} =\mathrm{21}\:\:\wedge\:\:^{\frac{{x}+{y}}{\mathrm{2}}} \mathrm{P}_{\mathrm{2}} =\mathrm{30} \\ $$$${Let}\:\:{x}+\frac{{y}}{\mathrm{2}}={n},\:\:^{{n}} \mathrm{C}_{\mathrm{2}} =\mathrm{21} \\ $$$$\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}=\mathrm{21} \\ $$$$\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{1}×\mathrm{2}}=\mathrm{21} \\ $$$${n}^{\mathrm{2}} −{n}=\mathrm{42} \\ $$$${n}^{\mathrm{2}} −{n}−\mathrm{42}=\mathrm{0} \\ $$$$\left({n}−\mathrm{7}\right)\left({n}+\mathrm{6}\right)=\mathrm{0} \\ $$$${n}=\mathrm{7}\:\:\:\mid\:\:{n}=−\mathrm{6}\left({discardable}\right) \\ $$$${x}+\frac{{y}}{\mathrm{2}}=\mathrm{7}\Rightarrow\mathrm{2}{x}+{y}=\mathrm{14}\:\:\:\:\:\left({i}\right) \\ $$$$−\:\:\:−\:\:−\:\:−\:\:− \\ $$$${Suppose}\:\frac{{x}+{y}}{\mathrm{2}}={m} \\ $$$$\:^{\frac{{x}+{y}}{\mathrm{2}}} \mathrm{P}_{\mathrm{2}} =\mathrm{30}\Rightarrow^{{m}} \mathrm{P}_{\mathrm{2}} =\mathrm{30}\:\:\: \\ $$$$\:\frac{{m}!}{\left({m}−\mathrm{2}\right)!}=\mathrm{30} \\ $$$${m}\left({m}−\mathrm{1}\right)=\mathrm{30} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{30}=\mathrm{0} \\ $$$$\left({m}−\mathrm{6}\right)\left({m}+\mathrm{5}\right)=\mathrm{0} \\ $$$${m}=\mathrm{6}\:\:\mid\:\:\:{m}=−\mathrm{5}\left({dicardable}\right) \\ $$$$\frac{{x}+{y}}{\mathrm{2}}=\mathrm{6}\Rightarrow{x}+{y}=\mathrm{12}\:\:\:\:\:\:\:\:\:\:\left({ii}\right) \\ $$$${From}\left({i}\right)\:\&\:\left({ii}\right): \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{10} \\ $$
Commented by sanusihammed last updated on 13/Jun/16
$${Thanks}\:{very}\:{much} \\ $$