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Question Number 34877 by NECx last updated on 12/May/18
4 couples are to take a photograph  with a newly wedded couple in a  wedding party.In how many ways  can this be done if:  i)the celebrated couple must stand  in the middle  ii)each couple must stand next to  each other  iii)the celebrated couple must not  stand next to each other
$$\mathrm{4}\:{couples}\:{are}\:{to}\:{take}\:{a}\:{photograph} \\ $$$${with}\:{a}\:{newly}\:{wedded}\:{couple}\:{in}\:{a} \\ $$$${wedding}\:{party}.{In}\:{how}\:{many}\:{ways} \\ $$$${can}\:{this}\:{be}\:{done}\:{if}: \\ $$$$\left.{i}\right){the}\:{celebrated}\:{couple}\:{must}\:{stand} \\ $$$${in}\:{the}\:{middle} \\ $$$$\left.{ii}\right){each}\:{couple}\:{must}\:{stand}\:{next}\:{to} \\ $$$${each}\:{other} \\ $$$$\left.{iii}\right){the}\:{celebrated}\:{couple}\:{must}\:{not} \\ $$$${stand}\:{next}\:{to}\:{each}\:{other} \\ $$
Commented by NECx last updated on 12/May/18
I′m so sorry I made typo error.Its  meant to be 4 couples not A couple.  It has been editdd.
$${I}'{m}\:{so}\:{sorry}\:{I}\:{made}\:{typo}\:{error}.{Its} \\ $$$${meant}\:{to}\:{be}\:\mathrm{4}\:{couples}\:{not}\:{A}\:{couple}. \\ $$$${It}\:{has}\:{been}\:{editdd}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/May/18
let new celebrated couple H_n ,W_n   old couple H_o ,W_o   1)H_o ,H_n ,W_n ,W_o      H_o ,W_n ,H_n ,W_o     W_o ,W_n ,H_n ,H_o      W_o ,H_n ,W_n ,H_o   four place to be filled...two extreme by 2 ways  and middle two place by two new couple  so 2×2=4 ways  2)H_n W_n H_o W_o    H_n W_n W_o H_o   W_n H_n H_o W_o   W_n H_n W_o H_o   H_o W_o H_n W_n   H_o W_o W_n H_n   W_o H_o H_n W_n   W_o H_o W_n H_n   so 8ways  3)the extreme two end place can be filled by either   new husband or new wife so 2 ways  middle two place by 2 ways so total 2×2=4 ways
$${let}\:{new}\:{celebrated}\:{couple}\:{H}_{{n}} ,{W}_{{n}} \\ $$$${old}\:{couple}\:{H}_{{o}} ,{W}_{{o}} \\ $$$$\left.\mathrm{1}\right){H}_{{o}} ,{H}_{{n}} ,{W}_{{n}} ,{W}_{{o}} \\ $$$$\:\:\:{H}_{{o}} ,{W}_{{n}} ,{H}_{{n}} ,{W}_{{o}} \\ $$$$\:\:{W}_{{o}} ,{W}_{{n}} ,{H}_{{n}} ,{H}_{{o}} \\ $$$$\:\:\:{W}_{{o}} ,{H}_{{n}} ,{W}_{{n}} ,{H}_{{o}} \\ $$$${four}\:{place}\:{to}\:{be}\:{filled}…{two}\:{extreme}\:{by}\:\mathrm{2}\:{ways} \\ $$$${and}\:{middle}\:{two}\:{place}\:{by}\:{two}\:{new}\:{couple} \\ $$$${so}\:\mathrm{2}×\mathrm{2}=\mathrm{4}\:{ways} \\ $$$$\left.\mathrm{2}\right){H}_{{n}} {W}_{{n}} {H}_{{o}} {W}_{{o}} \\ $$$$\:{H}_{{n}} {W}_{{n}} {W}_{{o}} {H}_{{o}} \\ $$$${W}_{{n}} {H}_{{n}} {H}_{{o}} {W}_{{o}} \\ $$$${W}_{{n}} {H}_{{n}} {W}_{{o}} {H}_{{o}} \\ $$$${H}_{{o}} {W}_{{o}} {H}_{{n}} {W}_{{n}} \\ $$$${H}_{{o}} {W}_{{o}} {W}_{{n}} {H}_{{n}} \\ $$$${W}_{{o}} {H}_{{o}} {H}_{{n}} {W}_{{n}} \\ $$$${W}_{{o}} {H}_{{o}} {W}_{{n}} {H}_{{n}} \\ $$$${so}\:\mathrm{8}{ways} \\ $$$$\left.\mathrm{3}\right){the}\:{extreme}\:{two}\:{end}\:{place}\:{can}\:{be}\:{filled}\:{by}\:{either}\: \\ $$$${new}\:{husband}\:{or}\:{new}\:{wife}\:{so}\:\mathrm{2}\:{ways} \\ $$$${middle}\:{two}\:{place}\:{by}\:\mathrm{2}\:{ways}\:{so}\:{total}\:\mathrm{2}×\mathrm{2}=\mathrm{4}\:{ways} \\ $$$$\:\:\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/May/18
1)total member in the party=4×2+2=10  1 2 3 4 5  6 7 8 9 10  there are ten seat to be occupied by party membdr  new couple to occupy seat number 5 and6s  remaining eight seat to be occupied by other  four couple  so 2!×8!  2)imagine each couple are attached together  so one couple maybe imagined one unit  so required permutation is 5!×2!  3)let celebrated couple are fabiquicked with  each other so that they can not be separated  the permutation so tbat new couple always  together is (4×2+1)!×2!  =9!×2!  so required permutation when new couple  never together is 10!−9!×2!
$$\left.\mathrm{1}\right){total}\:{member}\:{in}\:{the}\:{party}=\mathrm{4}×\mathrm{2}+\mathrm{2}=\mathrm{10} \\ $$$$\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\:\mathrm{6}\:\mathrm{7}\:\mathrm{8}\:\mathrm{9}\:\mathrm{10} \\ $$$${there}\:{are}\:{ten}\:{seat}\:{to}\:{be}\:{occupied}\:{by}\:{party}\:{membdr} \\ $$$${new}\:{couple}\:{to}\:{occupy}\:{seat}\:{number}\:\mathrm{5}\:{and}\mathrm{6}{s} \\ $$$${remaining}\:{eight}\:{seat}\:{to}\:{be}\:{occupied}\:{by}\:{other} \\ $$$${four}\:{couple} \\ $$$${so}\:\mathrm{2}!×\mathrm{8}! \\ $$$$\left.\mathrm{2}\right){imagine}\:{each}\:{couple}\:{are}\:{attached}\:{together} \\ $$$${so}\:{one}\:{couple}\:{maybe}\:{imagined}\:{one}\:{unit} \\ $$$${so}\:{required}\:{permutation}\:{is}\:\mathrm{5}!×\mathrm{2}! \\ $$$$\left.\mathrm{3}\right){let}\:{celebrated}\:{couple}\:{are}\:{fabiquicked}\:{with} \\ $$$${each}\:{other}\:{so}\:{that}\:{they}\:{can}\:{not}\:{be}\:{separated} \\ $$$${the}\:{permutation}\:{so}\:{tbat}\:{new}\:{couple}\:{always} \\ $$$${together}\:{is}\:\left(\mathrm{4}×\mathrm{2}+\mathrm{1}\right)!×\mathrm{2}! \\ $$$$=\mathrm{9}!×\mathrm{2}! \\ $$$${so}\:{required}\:{permutation}\:{when}\:{new}\:{couple} \\ $$$${never}\:{together}\:{is}\:\mathrm{10}!−\mathrm{9}!×\mathrm{2}! \\ $$
Commented by NECx last updated on 12/May/18
Thank you so much
$${Thank}\:{you}\:{so}\:{much} \\ $$
Commented by NECx last updated on 13/May/18
its really explanatory
$${its}\:{really}\:{explanatory} \\ $$

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