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Question Number 116397 by john santu last updated on 03/Oct/20
The number 1,2,3,4,...,7 are randomly  divided into two non −empty subsets.  The probability that the sum of the  numbers in the two subsets being  equal is (r/s) expressed in the lowest  term. Find r+s ?
$${The}\:{number}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},…,\mathrm{7}\:{are}\:{randomly} \\ $$$${divided}\:{into}\:{two}\:{non}\:−{empty}\:{subsets}. \\ $$$${The}\:{probability}\:{that}\:{the}\:{sum}\:{of}\:{the} \\ $$$${numbers}\:{in}\:{the}\:{two}\:{subsets}\:{being} \\ $$$${equal}\:{is}\:\frac{{r}}{{s}}\:{expressed}\:{in}\:{the}\:{lowest} \\ $$$${term}.\:{Find}\:{r}+{s}\:? \\ $$
Answered by mr W last updated on 03/Oct/20
1+2+3+4+5+6+7=28  each subset should have the sum 14.    subsets with 1 number & 6 numbers:  n=C_1 ^7 =7 possibilities  n_S =0  subsets with 2 numbers & 5 numbers:  n=C_2 ^7 =21 possibilities  n_S =0  subsets with 3 numbers & 4 numbers:  n=C_3 ^7 =35 possibilities  7+3+4, 7+2+5, 7+1+6, 6+5+3=14  n_S =4    p=((0+0+4)/(7+21+35))=(4/(63))=(r/s)  ⇒r+s=4+63=67    we can also get the total number of  possibilities via {_2 ^7 }=63.
$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}=\mathrm{28} \\ $$$${each}\:{subset}\:{should}\:{have}\:{the}\:{sum}\:\mathrm{14}. \\ $$$$ \\ $$$${subsets}\:{with}\:\mathrm{1}\:{number}\:\&\:\mathrm{6}\:{numbers}: \\ $$$${n}={C}_{\mathrm{1}} ^{\mathrm{7}} =\mathrm{7}\:{possibilities} \\ $$$${n}_{{S}} =\mathrm{0} \\ $$$${subsets}\:{with}\:\mathrm{2}\:{numbers}\:\&\:\mathrm{5}\:{numbers}: \\ $$$${n}={C}_{\mathrm{2}} ^{\mathrm{7}} =\mathrm{21}\:{possibilities} \\ $$$${n}_{{S}} =\mathrm{0} \\ $$$${subsets}\:{with}\:\mathrm{3}\:{numbers}\:\&\:\mathrm{4}\:{numbers}: \\ $$$${n}={C}_{\mathrm{3}} ^{\mathrm{7}} =\mathrm{35}\:{possibilities} \\ $$$$\mathrm{7}+\mathrm{3}+\mathrm{4},\:\mathrm{7}+\mathrm{2}+\mathrm{5},\:\mathrm{7}+\mathrm{1}+\mathrm{6},\:\mathrm{6}+\mathrm{5}+\mathrm{3}=\mathrm{14} \\ $$$${n}_{{S}} =\mathrm{4} \\ $$$$ \\ $$$${p}=\frac{\mathrm{0}+\mathrm{0}+\mathrm{4}}{\mathrm{7}+\mathrm{21}+\mathrm{35}}=\frac{\mathrm{4}}{\mathrm{63}}=\frac{{r}}{{s}} \\ $$$$\Rightarrow{r}+{s}=\mathrm{4}+\mathrm{63}=\mathrm{67} \\ $$$$ \\ $$$${we}\:{can}\:{also}\:{get}\:{the}\:{total}\:{number}\:{of} \\ $$$${possibilities}\:{via}\:\left\{_{\mathrm{2}} ^{\mathrm{7}} \right\}=\mathrm{63}. \\ $$
Commented by john santu last updated on 04/Oct/20
yes..thank you
$${yes}..{thank}\:{you} \\ $$
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 ((2^7 −2)/2) = 63. Note that since  1+2+3+...+7=28 ,the sum of the   number in each of the two groups is 14.  Note also that numbers 5,6,7 cannot  be in the same group since 5+6+7=18>14  case(1) only 6 &7 in the same group and   5 the other group. {2,3,4,5},{1,6,7}  case(2) only 5 &6 in the same group and  7 in the other group . {1,2,5,6}, {3,4,7}   {3,5,6}, { 1,2,4,7 }  case(3) only 5 &7 in the same group and  6 in the other group . {2,5,7}, {1,3,4,6}  Hence there are 4 such possibilities   Thus the required probability is  (4/(63)) yielding that p+q = 67
$${The}\:{total}\:{number}\:{of}\:{ways}\:{of}\:{dividing}\:\: \\ $$$${the}\:{seven}\:{numbers}\:{into}\:{two}\:{non}−{empty} \\ $$$${subsets}\:{is}\:\frac{\mathrm{2}^{\mathrm{7}} −\mathrm{2}}{\mathrm{2}}\:=\:\mathrm{63}.\:{Note}\:{that}\:{since} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{7}=\mathrm{28}\:,{the}\:{sum}\:{of}\:{the}\: \\ $$$${number}\:{in}\:{each}\:{of}\:{the}\:{two}\:{groups}\:{is}\:\mathrm{14}. \\ $$$${Note}\:{also}\:{that}\:{numbers}\:\mathrm{5},\mathrm{6},\mathrm{7}\:{cannot} \\ $$$${be}\:{in}\:{the}\:{same}\:{group}\:{since}\:\mathrm{5}+\mathrm{6}+\mathrm{7}=\mathrm{18}>\mathrm{14} \\ $$$${case}\left(\mathrm{1}\right)\:{only}\:\mathrm{6}\:\&\mathrm{7}\:{in}\:{the}\:{same}\:{group}\:{and}\: \\ $$$$\mathrm{5}\:{the}\:{other}\:{group}.\:\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\left\{\mathrm{1},\mathrm{6},\mathrm{7}\right\} \\ $$$${case}\left(\mathrm{2}\right)\:{only}\:\mathrm{5}\:\&\mathrm{6}\:{in}\:{the}\:{same}\:{group}\:{and} \\ $$$$\mathrm{7}\:{in}\:{the}\:{other}\:{group}\:.\:\left\{\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{6}\right\},\:\left\{\mathrm{3},\mathrm{4},\mathrm{7}\right\} \\ $$$$\:\left\{\mathrm{3},\mathrm{5},\mathrm{6}\right\},\:\left\{\:\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\:\right\} \\ $$$${case}\left(\mathrm{3}\right)\:{only}\:\mathrm{5}\:\&\mathrm{7}\:{in}\:{the}\:{same}\:{group}\:{and} \\ $$$$\mathrm{6}\:{in}\:{the}\:{other}\:{group}\:.\:\left\{\mathrm{2},\mathrm{5},\mathrm{7}\right\},\:\left\{\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{6}\right\} \\ $$$${Hence}\:{there}\:{are}\:\mathrm{4}\:{such}\:{possibilities}\: \\ $$$${Thus}\:{the}\:{required}\:{probability}\:{is} \\ $$$$\frac{\mathrm{4}}{\mathrm{63}}\:{yielding}\:{that}\:{p}+{q}\:=\:\mathrm{67} \\ $$

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