Multiple choice

There are 12 different balls in an urn . Number of ways in which 3 balls can by drawn from the urn four times in succession without replacement is equal to .

$\dfrac { 12! }{ (3!)^{ 4 }4! } $
$\dfrac { 12! }{ (3!)^{ 4 } } $
$\dfrac { 12! }{ { 9 }^{ 2 },4^{ 4 } } $
$\dfrac { 12! }{ (4!)^{ 3 } } $

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