Question Number 116397 by john santu last updated on 03/Oct/20
$${The}\:{number}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…,\mathrm{7}\:{are}\:{randomly} \\ $$$${divided}\:{into}\:{two}\:{non}\:−{empty}\:{subsets}. \\ $$$${The}\:{probability}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${numbers}\:{in}\:{the}\:{two}\:{subsets}\:{being} \\ $$$${equal}\:{is}\:\frac{{r}}{{s}}\:{expressed}\:{in}\:{the}\:{lowest} \\ $$$${term}.\:{Find}\:{r}+{s}\:? \\ $$
Answered by mr W last updated on 03/Oct/20
$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}=\mathrm{28} \\ $$$${each}\:{subset}\:{should}\:{have}\:{the}\:{sum}\:\mathrm{14}. \\ $$$$ \\ $$$${subsets}\:{with}\:\mathrm{1}\:{number}\:\&\:\mathrm{6}\:{numbers}: \\ $$$${n}={C}_{\mathrm{1}} ^{\mathrm{7}} =\mathrm{7}\:{possibilities} \\ $$$${n}_{{S}} =\mathrm{0} \\ $$$${subsets}\:{with}\:\mathrm{2}\:{numbers}\:\&\:\mathrm{5}\:{numbers}: \\ $$$${n}={C}_{\mathrm{2}} ^{\mathrm{7}} =\mathrm{21}\:{possibilities} \\ $$$${n}_{{S}} =\mathrm{0} \\ $$$${subsets}\:{with}\:\mathrm{3}\:{numbers}\:\&\:\mathrm{4}\:{numbers}: \\ $$$${n}={C}_{\mathrm{3}} ^{\mathrm{7}} =\mathrm{35}\:{possibilities} \\ $$$$\mathrm{7}+\mathrm{3}+\mathrm{4},\:\mathrm{7}+\mathrm{2}+\mathrm{5},\:\mathrm{7}+\mathrm{1}+\mathrm{6},\:\mathrm{6}+\mathrm{5}+\mathrm{3}=\mathrm{14} \\ $$$${n}_{{S}} =\mathrm{4} \\ $$$$ \\ $$$${p}=\frac{\mathrm{0}+\mathrm{0}+\mathrm{4}}{\mathrm{7}+\mathrm{21}+\mathrm{35}}=\frac{\mathrm{4}}{\mathrm{63}}=\frac{{r}}{{s}} \\ $$$$\Rightarrow{r}+{s}=\mathrm{4}+\mathrm{63}=\mathrm{67} \\ $$$$ \\ $$$${we}\:{can}\:{also}\:{get}\:{the}\:{total}\:{number}\:{of} \\ $$$${possibilities}\:{via}\:\left\{_{\mathrm{2}} ^{\mathrm{7}} \right\}=\mathrm{63}. \\ $$
Commented by john santu last updated on 04/Oct/20
$${yes}..{thank}\:{you} \\ $$
Answered by john santu last updated on 04/Oct/20
$${The}\:{total}\:{number}\:{of}\:{ways}\:{of}\:{dividing}\:\: \\ $$$${the}\:{seven}\:{numbers}\:{into}\:{two}\:{non}−{empty} \\ $$$${subsets}\:{is}\:\frac{\mathrm{2}^{\mathrm{7}} −\mathrm{2}}{\mathrm{2}}\:=\:\mathrm{63}.\:{Note}\:{that}\:{since} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{7}=\mathrm{28}\:,{the}\:{sum}\:{of}\:{the}\: \\ $$$${number}\:{in}\:{each}\:{of}\:{the}\:{two}\:{groups}\:{is}\:\mathrm{14}. \\ $$$${Note}\:{also}\:{that}\:{numbers}\:\mathrm{5},\mathrm{6},\mathrm{7}\:{cannot} \\ $$$${be}\:{in}\:{the}\:{same}\:{group}\:{since}\:\mathrm{5}+\mathrm{6}+\mathrm{7}=\mathrm{18}>\mathrm{14} \\ $$$${case}\left(\mathrm{1}\right)\:{only}\:\mathrm{6}\:\&\mathrm{7}\:{in}\:{the}\:{same}\:{group}\:{and}\: \\ $$$$\mathrm{5}\:{the}\:{other}\:{group}.\:\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\left\{\mathrm{1},\mathrm{6},\mathrm{7}\right\} \\ $$$${case}\left(\mathrm{2}\right)\:{only}\:\mathrm{5}\:\&\mathrm{6}\:{in}\:{the}\:{same}\:{group}\:{and} \\ $$$$\mathrm{7}\:{in}\:{the}\:{other}\:{group}\:.\:\left\{\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{6}\right\},\:\left\{\mathrm{3},\mathrm{4},\mathrm{7}\right\} \\ $$$$\:\left\{\mathrm{3},\mathrm{5},\mathrm{6}\right\},\:\left\{\:\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\:\right\} \\ $$$${case}\left(\mathrm{3}\right)\:{only}\:\mathrm{5}\:\&\mathrm{7}\:{in}\:{the}\:{same}\:{group}\:{and} \\ $$$$\mathrm{6}\:{in}\:{the}\:{other}\:{group}\:.\:\left\{\mathrm{2},\mathrm{5},\mathrm{7}\right\},\:\left\{\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{6}\right\} \\ $$$${Hence}\:{there}\:{are}\:\mathrm{4}\:{such}\:{possibilities}\: \\ $$$${Thus}\:{the}\:{required}\:{probability}\:{is} \\ $$$$\frac{\mathrm{4}}{\mathrm{63}}\:{yielding}\:{that}\:{p}+{q}\:=\:\mathrm{67} \\ $$