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Question-6090




Question Number 6090 by sanusihammed last updated on 13/Jun/16
Answered by Rasheed Soomro last updated on 13/Jun/16
^(x+(y/2)) C_2 =21  ∧ ^((x+y)/2) P_2 =30  Let  x+(y/2)=n, ^n C_2 =21  ((n!)/(2!(n−2)!))=21  ((n(n−1))/(1×2))=21  n^2 −n=42  n^2 −n−42=0  (n−7)(n+6)=0  n=7   ∣  n=−6(discardable)  x+(y/2)=7⇒2x+y=14     (i)  −   −  −  −  −  Suppose ((x+y)/2)=m  ^((x+y)/2) P_2 =30⇒^m P_2 =30      ((m!)/((m−2)!))=30  m(m−1)=30  m^2 −m−30=0  (m−6)(m+5)=0  m=6  ∣   m=−5(dicardable)  ((x+y)/2)=6⇒x+y=12          (ii)  From(i) & (ii):  x=2, y=10
$$\:^{{x}+\frac{{y}}{\mathrm{2}}} \mathrm{C}_{\mathrm{2}} =\mathrm{21}\:\:\wedge\:\:^{\frac{{x}+{y}}{\mathrm{2}}} \mathrm{P}_{\mathrm{2}} =\mathrm{30} \\ $$$${Let}\:\:{x}+\frac{{y}}{\mathrm{2}}={n},\:\:^{{n}} \mathrm{C}_{\mathrm{2}} =\mathrm{21} \\ $$$$\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}=\mathrm{21} \\ $$$$\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{1}×\mathrm{2}}=\mathrm{21} \\ $$$${n}^{\mathrm{2}} −{n}=\mathrm{42} \\ $$$${n}^{\mathrm{2}} −{n}−\mathrm{42}=\mathrm{0} \\ $$$$\left({n}−\mathrm{7}\right)\left({n}+\mathrm{6}\right)=\mathrm{0} \\ $$$${n}=\mathrm{7}\:\:\:\mid\:\:{n}=−\mathrm{6}\left({discardable}\right) \\ $$$${x}+\frac{{y}}{\mathrm{2}}=\mathrm{7}\Rightarrow\mathrm{2}{x}+{y}=\mathrm{14}\:\:\:\:\:\left({i}\right) \\ $$$$−\:\:\:−\:\:−\:\:−\:\:− \\ $$$${Suppose}\:\frac{{x}+{y}}{\mathrm{2}}={m} \\ $$$$\:^{\frac{{x}+{y}}{\mathrm{2}}} \mathrm{P}_{\mathrm{2}} =\mathrm{30}\Rightarrow^{{m}} \mathrm{P}_{\mathrm{2}} =\mathrm{30}\:\:\: \\ $$$$\:\frac{{m}!}{\left({m}−\mathrm{2}\right)!}=\mathrm{30} \\ $$$${m}\left({m}−\mathrm{1}\right)=\mathrm{30} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{30}=\mathrm{0} \\ $$$$\left({m}−\mathrm{6}\right)\left({m}+\mathrm{5}\right)=\mathrm{0} \\ $$$${m}=\mathrm{6}\:\:\mid\:\:\:{m}=−\mathrm{5}\left({dicardable}\right) \\ $$$$\frac{{x}+{y}}{\mathrm{2}}=\mathrm{6}\Rightarrow{x}+{y}=\mathrm{12}\:\:\:\:\:\:\:\:\:\:\left({ii}\right) \\ $$$${From}\left({i}\right)\:\&\:\left({ii}\right): \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{10} \\ $$
Commented by sanusihammed last updated on 13/Jun/16
Thanks very much
$${Thanks}\:{very}\:{much} \\ $$

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