The correct option is D None of these
When 3dicearethrownsimultaneouslythesumwhichisaperfectsquarecanbe4,9,164canbeformedfrom:(1,1,2)9canbeformedfrom:(1,2,6), (1,3,5), (1,4,4), (2,2,5), (2,3,4), (3,3,3)and16canbeformedfrom:(6,6,4), (6,5,5)Let′scalltheeventofgettingsum,whichisaperfectsquare,EThenP\left( E \right) =\left( 31 \right) P\left( 1 \right) P\left( 1 \right) P\left( 2 \right) +\left( 3! \right) \left\{ P\left( 1 \right) P\left( 2 \right) P\left( 6 \right) +P\left( 1 \right) P\left( 3 \right) P\left( 5 \right) +P\left( 2 \right) P\left( 3 \right) P\left( 4 \right) \right\} \ +\left( 31 \right) \left\{ P\left( 1 \right) P\left( 4 \right) P\left( 4 \right) +P\left( 2 \right) P\left( 2 \right) P\left( 5 \right) \right\} +P\left( 3 \right) P\left( 3 \right) P\left( 3 \right) \ +\left( 31 \right) \left\{ P\left( 6 \right) P\left( 6 \right) P\left( 4 \right) +P\left( 6 \right) P\left( 5 \right) P\left( 5 \right) \right\} \ \Rightarrow P\left( E \right) =3\times \left( { \dfrac { 1^3 }{ 6^3 } }\right) +\left( 3! \right) \left\{ 3\times \left( { \dfrac { 1^3 }{ 6^3 } }\right) \right\} +3\times \left\{ 3\times \left( { \dfrac { 1^3 }{ 6^3 } } \right) \right\} +\left( { \dfrac { 1^3 }{ 6^3 } } \right) +3\times 2\times \left( { \dfrac { 1^3 }{ 6^3 } } \right) \ =\dfrac { 37 }{ 216 } $