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Question-178626




Question Number 178626 by mr W last updated on 19/Oct/22
Commented by Frix last updated on 20/Oct/22
am I right with this conception?  put 1 student of each college at a board and  then “draw” the others which is the same  as try each possible combination of the 10  remaining. we have 10! different combinations  and if in a certain combination at least one  pair of “friends” appears, this combination  is not allowed
$$\mathrm{am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{with}\:\mathrm{this}\:\mathrm{conception}? \\ $$$$\mathrm{put}\:\mathrm{1}\:\mathrm{student}\:\mathrm{of}\:\mathrm{each}\:\mathrm{college}\:\mathrm{at}\:\mathrm{a}\:\mathrm{board}\:\mathrm{and} \\ $$$$\mathrm{then}\:“\mathrm{draw}''\:\mathrm{the}\:\mathrm{others}\:\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{as}\:\mathrm{try}\:\mathrm{each}\:\mathrm{possible}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{the}\:\mathrm{10} \\ $$$$\mathrm{remaining}.\:\mathrm{we}\:\mathrm{have}\:\mathrm{10}!\:\mathrm{different}\:\mathrm{combinations} \\ $$$$\mathrm{and}\:\mathrm{if}\:\mathrm{in}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{combination}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one} \\ $$$$\mathrm{pair}\:\mathrm{of}\:“\mathrm{friends}''\:\mathrm{appears},\:\mathrm{this}\:\mathrm{combination} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{allowed} \\ $$
Commented by Frix last updated on 20/Oct/22
I wrote a simulation program and got  p≈36.82%±1.78%  with a confidence of 99.95% at n=9000  this would be (7/(19))
$$\mathrm{I}\:\mathrm{wrote}\:\mathrm{a}\:\mathrm{simulation}\:\mathrm{program}\:\mathrm{and}\:\mathrm{got} \\ $$$${p}\approx\mathrm{36}.\mathrm{82\%}\pm\mathrm{1}.\mathrm{78\%} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{confidence}\:\mathrm{of}\:\mathrm{99}.\mathrm{95\%}\:\mathrm{at}\:{n}=\mathrm{9000} \\ $$$$\mathrm{this}\:\mathrm{would}\:\mathrm{be}\:\frac{\mathrm{7}}{\mathrm{19}} \\ $$
Commented by MJS_new last updated on 24/Oct/22
it′s not clear...  example with only 4 students from 2 colleges  (1) drawing all possible “lines” of students  leads to 24 different combinations of which  8 are not allowed ⇒ p=(2/3)  (2) student a_1  and b_1  take seat at 2 different  tables ⇒ 2 possibilities for the others of which  1 is not allowed ⇒ p=(1/2)
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{clear}… \\ $$$$\mathrm{example}\:\mathrm{with}\:\mathrm{only}\:\mathrm{4}\:\mathrm{students}\:\mathrm{from}\:\mathrm{2}\:\mathrm{colleges} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{drawing}\:\mathrm{all}\:\mathrm{possible}\:“\mathrm{lines}''\:\mathrm{of}\:\mathrm{students} \\ $$$$\mathrm{leads}\:\mathrm{to}\:\mathrm{24}\:\mathrm{different}\:\mathrm{combinations}\:\mathrm{of}\:\mathrm{which} \\ $$$$\mathrm{8}\:\mathrm{are}\:\mathrm{not}\:\mathrm{allowed}\:\Rightarrow\:{p}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{student}\:{a}_{\mathrm{1}} \:\mathrm{and}\:{b}_{\mathrm{1}} \:\mathrm{take}\:\mathrm{seat}\:\mathrm{at}\:\mathrm{2}\:\mathrm{different} \\ $$$$\mathrm{tables}\:\Rightarrow\:\mathrm{2}\:\mathrm{possibilities}\:\mathrm{for}\:\mathrm{the}\:\mathrm{others}\:\mathrm{of}\:\mathrm{which} \\ $$$$\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{allowed}\:\Rightarrow\:{p}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Acem last updated on 19/Oct/22
  P_(2p fr.sm.colg.not play) = ((18)/(19))
$$ \\ $$$${P}_{\mathrm{2}{p}\:{fr}.{sm}.{colg}.{not}\:{play}} =\:\frac{\mathrm{18}}{\mathrm{19}} \\ $$
Commented by Acem last updated on 19/Oct/22
That′s as i understood that there are only 10   matches.
$${That}'{s}\:{as}\:{i}\:{understood}\:{that}\:{there}\:{are}\:{only}\:\mathrm{10} \\ $$$$\:{matches}. \\ $$
Commented by Frix last updated on 19/Oct/22
I think it should be around 33%, maybe  a bit more but ((18)/(19))≈94.74% seems too much
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{around}\:\mathrm{33\%},\:\mathrm{maybe} \\ $$$$\mathrm{a}\:\mathrm{bit}\:\mathrm{more}\:\mathrm{but}\:\frac{\mathrm{18}}{\mathrm{19}}\approx\mathrm{94}.\mathrm{74\%}\:\mathrm{seems}\:\mathrm{too}\:\mathrm{much} \\ $$
Commented by Acem last updated on 19/Oct/22
A game between two students from same college   must be low in front of each one of them   may plays with someone of other 18 students
$${A}\:{game}\:{between}\:{two}\:{students}\:{from}\:{same}\:{college} \\ $$$$\:{must}\:{be}\:{low}\:{in}\:{front}\:{of}\:{each}\:{one}\:{of}\:{them} \\ $$$$\:{may}\:{plays}\:{with}\:{someone}\:{of}\:{other}\:\mathrm{18}\:{students} \\ $$
Commented by mr W last updated on 19/Oct/22
i think the question is misunderstood  by you.  to divide the 20 students into 10 pairs,  there are totally n_1  ways. to divide  them into 10 pairs such that no pair   contains students from the same  college there are n_2  ways. now we  should find n_1  and n_2  to get p=(n_2 /n_1 ).
$${i}\:{think}\:{the}\:{question}\:{is}\:{misunderstood} \\ $$$${by}\:{you}. \\ $$$${to}\:{divide}\:{the}\:\mathrm{20}\:{students}\:{into}\:\mathrm{10}\:{pairs}, \\ $$$${there}\:{are}\:{totally}\:{n}_{\mathrm{1}} \:{ways}.\:{to}\:{divide} \\ $$$${them}\:{into}\:\mathrm{10}\:{pairs}\:{such}\:{that}\:{no}\:{pair}\: \\ $$$${contains}\:{students}\:{from}\:{the}\:{same} \\ $$$${college}\:{there}\:{are}\:{n}_{\mathrm{2}} \:{ways}.\:{now}\:{we} \\ $$$${should}\:{find}\:{n}_{\mathrm{1}} \:{and}\:{n}_{\mathrm{2}} \:{to}\:{get}\:{p}=\frac{{n}_{\mathrm{2}} }{{n}_{\mathrm{1}} }. \\ $$
Commented by Acem last updated on 20/Oct/22
Yes i know that as well
$${Yes}\:{i}\:{know}\:{that}\:{as}\:{well} \\ $$
Commented by mr W last updated on 20/Oct/22
what did you get for n_1  and n_2 ?
$${what}\:{did}\:{you}\:{get}\:{for}\:{n}_{\mathrm{1}} \:{and}\:{n}_{\mathrm{2}} ? \\ $$
Commented by Acem last updated on 20/Oct/22
190 paires represent n_1  , ten out of it represent   num.of paires of students from same college   hence 180 represent n_2
$$\mathrm{190}\:{paires}\:{represent}\:{n}_{\mathrm{1}} \:,\:{ten}\:{out}\:{of}\:{it}\:{represent} \\ $$$$\:{num}.{of}\:{paires}\:{of}\:{students}\:{from}\:{same}\:{college} \\ $$$$\:{hence}\:\mathrm{180}\:{represent}\:{n}_{\mathrm{2}} \\ $$
Commented by mr W last updated on 20/Oct/22
to divide 20 people into 10 groups  such that each group has exactly 2  people there are totally  n_1 =((20!)/((2!)^(10) 10!))=654 729 075 ways.
$${to}\:{divide}\:\mathrm{20}\:{people}\:{into}\:\mathrm{10}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{exactly}\:\mathrm{2} \\ $$$${people}\:{there}\:{are}\:{totally} \\ $$$${n}_{\mathrm{1}} =\frac{\mathrm{20}!}{\left(\mathrm{2}!\right)^{\mathrm{10}} \mathrm{10}!}=\mathrm{654}\:\mathrm{729}\:\mathrm{075}\:{ways}. \\ $$
Commented by Acem last updated on 20/Oct/22
You are correct about this, but what′s the point   of dividing them into these groups by this method   as long as every student may plays against only   one of 19 other opponents with exception of   repetition. At that time we get  190 possible   paires as a sample space.
$${You}\:{are}\:{correct}\:{about}\:{this},\:{but}\:{what}'{s}\:{the}\:{point} \\ $$$$\:{of}\:{dividing}\:{them}\:{into}\:{these}\:{groups}\:{by}\:{this}\:{method} \\ $$$$\:{as}\:{long}\:{as}\:{every}\:{student}\:{may}\:{plays}\:{against}\:{only} \\ $$$$\:{one}\:{of}\:\mathrm{19}\:{other}\:{opponents}\:{with}\:{exception}\:{of} \\ $$$$\:{repetition}.\:{At}\:{that}\:{time}\:{we}\:{get}\:\:\mathrm{190}\:{possible} \\ $$$$\:{paires}\:{as}\:{a}\:{sample}\:{space}. \\ $$$$ \\ $$
Commented by Acem last updated on 20/Oct/22
Give me a sample of some elements of the    sample space other the {(Jack_(col.1) , Toni_(col.8) ),   (Jack_(col1) , Smith_(col.3) ).... (Sam_(col.1) , Jeorge_(col.5) )...,   (Jack_(col.1) , Sam_(col.1) ).....}   and number of these pairses are 190
$${Give}\:{me}\:{a}\:{sample}\:{of}\:{some}\:{elements}\:{of}\:{the}\: \\ $$$$\:{sample}\:{space}\:{other}\:{the}\:\left\{\left({Jack}_{{col}.\mathrm{1}} ,\:{Toni}_{{col}.\mathrm{8}} \right),\right. \\ $$$$\:\left({Jack}_{{col}\mathrm{1}} ,\:{Smith}_{{col}.\mathrm{3}} \right)….\:\left({Sam}_{{col}.\mathrm{1}} ,\:{Jeorge}_{{col}.\mathrm{5}} \right)…, \\ $$$$\left.\:\left({Jack}_{{col}.\mathrm{1}} ,\:{Sam}_{{col}.\mathrm{1}} \right)…..\right\} \\ $$$$\:{and}\:{number}\:{of}\:{these}\:{pairses}\:{are}\:\mathrm{190} \\ $$$$ \\ $$
Commented by mr W last updated on 20/Oct/22
can you calculate n_2 , the number of  ways to divide the 20 people into 10  groups such that any two peoplefrom  the same college are not in the same  group?
$${can}\:{you}\:{calculate}\:{n}_{\mathrm{2}} ,\:{the}\:{number}\:{of} \\ $$$${ways}\:{to}\:{divide}\:{the}\:\mathrm{20}\:{people}\:{into}\:\mathrm{10} \\ $$$${groups}\:{such}\:{that}\:{any}\:{two}\:{peoplefrom} \\ $$$${the}\:{same}\:{college}\:{are}\:{not}\:{in}\:{the}\:{same} \\ $$$${group}? \\ $$
Commented by Acem last updated on 21/Oct/22
Forming paires:   10_(10 Col.) ^(ways) ×2∣×9_(9 col.) ^(w.) ×2  ⇒ 1st_P  (Jack, Toni)     2_(2Col.) ^(ways) ×1∣×1_(1 col.) ^(w.)  +^(OR)  2_(2Col.) ^(ways) ×1∣×9_(1 col.) ^(w.) ×1 +^(OR)    8_(frm 8 Col.) ^(ways) ×2∣×7_(frm 7 col.) ^(w.) ×2 ⇒ 2nd_P  (Sam, Dani)     2_(frm 2Col.) ^(ways) ×1∣×1_(frm 1 col.) ^(w.)  +^(OR)  complex ways +^(OR)    6_(frm 6 Col.) ^(ways) ×2∣×5_(frm 5 col.) ^(w.) ×2 ⇒ 3rd_P  (You, Me)  ⋮  ⋮  The tenth pairse:   2_(2Col.) ^(here i think 1 way not 2) ×1∣_(9th Col.) ×1_(1 col.) ^(w.) ∣_(10th Col)                                ⇒ 10th_P  (Tom, Paskal)   n_2 = Sum above     That′s a high hard work, and i don′t know why    Hey students shake hands with each other!    And i get 190 paires belong to the Sample Space     Ten out of it represent 2stu.frm.sm.Col.
$${Forming}\:{paires}: \\ $$$$\:\mathrm{10}_{\mathrm{10}\:{Col}.} ^{{ways}} ×\mathrm{2}\mid×\mathrm{9}_{\mathrm{9}\:{col}.} ^{{w}.} ×\mathrm{2}\:\:\Rightarrow\:\mathrm{1}{st}_{{P}} \:\left({Jack},\:{Toni}\right) \\ $$$$ \\ $$$$\:\mathrm{2}_{\mathrm{2}{Col}.} ^{{ways}} ×\mathrm{1}\mid×\mathrm{1}_{\mathrm{1}\:{col}.} ^{{w}.} \:\overset{{OR}} {+}\:\mathrm{2}_{\mathrm{2}{Col}.} ^{{ways}} ×\mathrm{1}\mid×\mathrm{9}_{\mathrm{1}\:{col}.} ^{{w}.} ×\mathrm{1}\:\overset{{OR}} {+} \\ $$$$\:\mathrm{8}_{{frm}\:\mathrm{8}\:{Col}.} ^{{ways}} ×\mathrm{2}\mid×\mathrm{7}_{{frm}\:\mathrm{7}\:{col}.} ^{{w}.} ×\mathrm{2}\:\Rightarrow\:\mathrm{2}{nd}_{{P}} \:\left({Sam},\:{Dani}\right) \\ $$$$ \\ $$$$\:\mathrm{2}_{{frm}\:\mathrm{2}{Col}.} ^{{ways}} ×\mathrm{1}\mid×\mathrm{1}_{{frm}\:\mathrm{1}\:{col}.} ^{{w}.} \:\overset{{OR}} {+}\:{complex}\:{ways}\:\overset{{OR}} {+} \\ $$$$\:\mathrm{6}_{{frm}\:\mathrm{6}\:{Col}.} ^{{ways}} ×\mathrm{2}\mid×\mathrm{5}_{{frm}\:\mathrm{5}\:{col}.} ^{{w}.} ×\mathrm{2}\:\Rightarrow\:\mathrm{3}{rd}_{{P}} \:\left({You},\:{Me}\right) \\ $$$$\vdots \\ $$$$\vdots \\ $$$${The}\:{tenth}\:{pairse}: \\ $$$$\:\mathrm{2}_{\mathrm{2}{Col}.} ^{{here}\:{i}\:{think}\:\mathrm{1}\:{way}\:{not}\:\mathrm{2}} ×\mathrm{1}\mid_{\mathrm{9}{th}\:{Col}.} ×\mathrm{1}_{\mathrm{1}\:{col}.} ^{{w}.} \mid_{\mathrm{10}{th}\:{Col}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{10}{th}_{{P}} \:\left({Tom},\:{Paskal}\right) \\ $$$$\:{n}_{\mathrm{2}} =\:{Sum}\:{above} \\ $$$$\: \\ $$$${That}'{s}\:{a}\:{high}\:{hard}\:{work},\:{and}\:{i}\:{don}'{t}\:{know}\:{why} \\ $$$$ \\ $$$$\boldsymbol{{Hey}}\:\boldsymbol{{students}}\:{shake}\:{hands}\:{with}\:{each}\:{other}! \\ $$$$ \\ $$$${And}\:{i}\:{get}\:\mathrm{190}\:{paires}\:{belong}\:{to}\:{the}\:\boldsymbol{{Sample}}\:\boldsymbol{{Space}} \\ $$$$ \\ $$$$\:{Ten}\:{out}\:{of}\:{it}\:{represent}\:\mathrm{2}{stu}.{frm}.{sm}.{Col}. \\ $$$$ \\ $$$$ \\ $$
Commented by Acem last updated on 21/Oct/22
  Can you give me a sample of some elements   of the  sample space other the {(Jack_(col.1) , Toni_(col.8) ),   (Jack_(col1) , Smith_(col.3) ).... (Sam_(col.1) , Jeorge_(col.5) )...,   (Jack_(col.1) , Sam_(col.1) ).....}   and number of these pairses are 190
$$ \\ $$$${Can}\:{you}\:{give}\:{me}\:{a}\:{sample}\:{of}\:{some}\:{elements} \\ $$$$\:{of}\:{the}\:\:{sample}\:{space}\:{other}\:{the}\:\left\{\left({Jack}_{{col}.\mathrm{1}} ,\:{Toni}_{{col}.\mathrm{8}} \right),\right. \\ $$$$\:\left({Jack}_{{col}\mathrm{1}} ,\:{Smith}_{{col}.\mathrm{3}} \right)….\:\left({Sam}_{{col}.\mathrm{1}} ,\:{Jeorge}_{{col}.\mathrm{5}} \right)…, \\ $$$$\left.\:\left({Jack}_{{col}.\mathrm{1}} ,\:{Sam}_{{col}.\mathrm{1}} \right)…..\right\} \\ $$$$\:{and}\:{number}\:{of}\:{these}\:{pairses}\:{are}\:\mathrm{190} \\ $$$$ \\ $$
Commented by Acem last updated on 21/Oct/22
This issue confused me. At first i thought just   like you, then i turned my attention to interresting   of  that each one can′t play but with one of 19 person,   therefore i formed the sample space above and it   is true.... thought it′s true.... it′s not true! and   the issue is not here, it′s about forming   10 mathces, so the sample space is for matches   not for people. Each item of this sample   space represent 10 games “10 pairses”.     What i′ve just said i think it′s correcrt, but really   this issue made me feel tired
$${This}\:{issue}\:{confused}\:{me}.\:{At}\:{first}\:{i}\:{thought}\:{just} \\ $$$$\:{like}\:{you},\:{then}\:{i}\:{turned}\:{my}\:{attention}\:{to}\:{interresting} \\ $$$$\:{of}\:\:{that}\:{each}\:{one}\:{can}'{t}\:{play}\:{but}\:{with}\:{one}\:{of}\:\mathrm{19}\:{person}, \\ $$$$\:{therefore}\:{i}\:{formed}\:{the}\:{sample}\:{space}\:{above}\:{and}\:{it} \\ $$$$\:{is}\:{true}….\:{thought}\:{it}'{s}\:{true}….\:{it}'{s}\:{not}\:{true}!\:{and} \\ $$$$\:{the}\:{issue}\:{is}\:{not}\:{here},\:{it}'{s}\:{about}\:{forming} \\ $$$$\:\mathrm{10}\:{mathces},\:{so}\:{the}\:{sample}\:{space}\:{is}\:{for}\:{matches} \\ $$$$\:{not}\:{for}\:{people}.\:{Each}\:{item}\:{of}\:{this}\:{sample} \\ $$$$\:{space}\:{represent}\:\mathrm{10}\:{games}\:“\mathrm{10}\:{pairses}''. \\ $$$$ \\ $$$$\:{What}\:{i}'{ve}\:{just}\:{said}\:{i}\:{think}\:{it}'{s}\:{correcrt},\:{but}\:{really} \\ $$$$\:{this}\:{issue}\:{made}\:{me}\:{feel}\:{tired} \\ $$
Commented by mr W last updated on 21/Oct/22
this is a very old question in the forum  which is not solved yet.
$${this}\:{is}\:{a}\:{very}\:{old}\:{question}\:{in}\:{the}\:{forum} \\ $$$${which}\:{is}\:{not}\:{solved}\:{yet}. \\ $$
Commented by mr W last updated on 21/Oct/22
the answer should be p≈0.55%.
$${the}\:{answer}\:{should}\:{be}\:{p}\approx\mathrm{0}.\mathrm{55\%}. \\ $$
Commented by MJS_new last updated on 24/Oct/22
where is this answer from?  I tried to figure how this develops.  with n=4 players we have  12 34 ×  13 24 ✓  14 23 ✓  each stands for 2^(n/2) ×(n/2)!=8 combinations  with n=6 players we have  12 34 56 ×  12 35 46 ×  12 36 45 ×  13 24 56 ×  13 25 46 ✓  13 26 45 ✓  14 23 56 ×  14 25 36 ✓  14 26 35 ✓  15 23 46 ✓  15 24 36 ✓  15 26 34 ×  16 23 45 ✓  16 24 35 ✓  16 25 34 ×  each stands for 2^(n/2) ×(n/2)!=48 combinations    I also did this for n=8  it seems to me to follow this rule:  n=4 N_4 =3 P_4 =1 p=(P_4 /N_4 )=(1/3)  n=6 N_6 =5N_4 =15 P_6 =N_4 (1+(5−1)/3)=7 p=(P_6 /N_6 )=(7/(15))  n=8 N_8 =7N_6 =105 P_8 =N_6 (1+(7−1)/3)=45 p=(P_8 /N_8 )=((45)/(105))=(3/7)  n=2k N_(2k) =(2k−1)N_(2k−2)  P_(2k) =N_(2k−2) (1+(2k−1)/3) p=(P_(2k) /N_(2k) )  maybe somebody can verify this but for  n=20 this gives (7/(19))
$$\mathrm{where}\:\mathrm{is}\:\mathrm{this}\:\mathrm{answer}\:\mathrm{from}? \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{figure}\:\mathrm{how}\:\mathrm{this}\:\mathrm{develops}. \\ $$$$\mathrm{with}\:{n}=\mathrm{4}\:\mathrm{players}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{12}\:\mathrm{34}\:× \\ $$$$\mathrm{13}\:\mathrm{24}\:\checkmark \\ $$$$\mathrm{14}\:\mathrm{23}\:\checkmark \\ $$$$\mathrm{each}\:\mathrm{stands}\:\mathrm{for}\:\mathrm{2}^{{n}/\mathrm{2}} ×\left({n}/\mathrm{2}\right)!=\mathrm{8}\:\mathrm{combinations} \\ $$$$\mathrm{with}\:{n}=\mathrm{6}\:\mathrm{players}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{12}\:\mathrm{34}\:\mathrm{56}\:× \\ $$$$\mathrm{12}\:\mathrm{35}\:\mathrm{46}\:× \\ $$$$\mathrm{12}\:\mathrm{36}\:\mathrm{45}\:× \\ $$$$\mathrm{13}\:\mathrm{24}\:\mathrm{56}\:× \\ $$$$\mathrm{13}\:\mathrm{25}\:\mathrm{46}\:\checkmark \\ $$$$\mathrm{13}\:\mathrm{26}\:\mathrm{45}\:\checkmark \\ $$$$\mathrm{14}\:\mathrm{23}\:\mathrm{56}\:× \\ $$$$\mathrm{14}\:\mathrm{25}\:\mathrm{36}\:\checkmark \\ $$$$\mathrm{14}\:\mathrm{26}\:\mathrm{35}\:\checkmark \\ $$$$\mathrm{15}\:\mathrm{23}\:\mathrm{46}\:\checkmark \\ $$$$\mathrm{15}\:\mathrm{24}\:\mathrm{36}\:\checkmark \\ $$$$\mathrm{15}\:\mathrm{26}\:\mathrm{34}\:× \\ $$$$\mathrm{16}\:\mathrm{23}\:\mathrm{45}\:\checkmark \\ $$$$\mathrm{16}\:\mathrm{24}\:\mathrm{35}\:\checkmark \\ $$$$\mathrm{16}\:\mathrm{25}\:\mathrm{34}\:× \\ $$$$\mathrm{each}\:\mathrm{stands}\:\mathrm{for}\:\mathrm{2}^{{n}/\mathrm{2}} ×\left({n}/\mathrm{2}\right)!=\mathrm{48}\:\mathrm{combinations} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{also}\:\mathrm{did}\:\mathrm{this}\:\mathrm{for}\:{n}=\mathrm{8} \\ $$$$\mathrm{it}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{me}\:\mathrm{to}\:\mathrm{follow}\:\mathrm{this}\:\mathrm{rule}: \\ $$$${n}=\mathrm{4}\:{N}_{\mathrm{4}} =\mathrm{3}\:{P}_{\mathrm{4}} =\mathrm{1}\:{p}=\frac{{P}_{\mathrm{4}} }{{N}_{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${n}=\mathrm{6}\:{N}_{\mathrm{6}} =\mathrm{5}{N}_{\mathrm{4}} =\mathrm{15}\:{P}_{\mathrm{6}} ={N}_{\mathrm{4}} \left(\mathrm{1}+\left(\mathrm{5}−\mathrm{1}\right)/\mathrm{3}\right)=\mathrm{7}\:{p}=\frac{{P}_{\mathrm{6}} }{{N}_{\mathrm{6}} }=\frac{\mathrm{7}}{\mathrm{15}} \\ $$$${n}=\mathrm{8}\:{N}_{\mathrm{8}} =\mathrm{7}{N}_{\mathrm{6}} =\mathrm{105}\:{P}_{\mathrm{8}} ={N}_{\mathrm{6}} \left(\mathrm{1}+\left(\mathrm{7}−\mathrm{1}\right)/\mathrm{3}\right)=\mathrm{45}\:{p}=\frac{{P}_{\mathrm{8}} }{{N}_{\mathrm{8}} }=\frac{\mathrm{45}}{\mathrm{105}}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$${n}=\mathrm{2}{k}\:{N}_{\mathrm{2}{k}} =\left(\mathrm{2}{k}−\mathrm{1}\right){N}_{\mathrm{2}{k}−\mathrm{2}} \:{P}_{\mathrm{2}{k}} ={N}_{\mathrm{2}{k}−\mathrm{2}} \left(\mathrm{1}+\left(\mathrm{2}{k}−\mathrm{1}\right)/\mathrm{3}\right)\:{p}=\frac{{P}_{\mathrm{2}{k}} }{{N}_{\mathrm{2}{k}} } \\ $$$$\mathrm{maybe}\:\mathrm{somebody}\:\mathrm{can}\:\mathrm{verify}\:\mathrm{this}\:\mathrm{but}\:\mathrm{for} \\ $$$${n}=\mathrm{20}\:\mathrm{this}\:\mathrm{gives}\:\frac{\mathrm{7}}{\mathrm{19}} \\ $$
Commented by Frix last updated on 24/Oct/22
but your p is the “wrong side”, you must  take 1−p
$$\mathrm{but}\:\mathrm{your}\:{p}\:\mathrm{is}\:\mathrm{the}\:“\mathrm{wrong}\:\mathrm{side}'',\:\mathrm{you}\:\mathrm{must} \\ $$$$\mathrm{take}\:\mathrm{1}−{p} \\ $$
Commented by MJS_new last updated on 24/Oct/22
yes you′re right
$$\mathrm{yes}\:\mathrm{you}'\mathrm{re}\:\mathrm{right} \\ $$

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