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