Question Number 97323 by mathocean1 last updated on 07/Jun/20
$${Mr}\:{Peter}\:{has}\:\mathrm{4}\:{children}.\:{x}\:{are}\:{in}\: \\ $$$${class}\:{C}\:{and}\:{y}\:{are}\:{in}\:{class}\:{D}.\:{x}\geqslant\mathrm{1}\:{and} \\ $$$${y}\geqslant\mathrm{1}.\:{Show}\:{that}\:{the}\:{number}\:{of}\:{possibility}\: \\ $$$${to}\:{choose}\:{at}\:{random}\:{and}\:{simultaneous} \\ $$$$\mathrm{2}\:{children}\:{in}\:{same}\:{class}\:{verify}\:{this} \\ $$$${equation}\:{p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6} \\ $$
Answered by mr W last updated on 07/Jun/20
$${x}\:{in}\:{class}\:{C},\:\mathrm{4}−{x}\:{in}\:{class}\:{D}. \\ $$$${p}={C}_{\mathrm{2}} ^{{x}} +{C}_{\mathrm{2}} ^{\mathrm{4}−{x}} =\frac{{x}\left({x}−\mathrm{1}\right)+\left(\mathrm{4}−{x}\right)\left(\mathrm{4}−{x}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6} \\ $$