Let α = ( 4 ! ) ! ( 4 ! ) 3 ! and β = ( 5 ! ) ! ( 5 ! ) 4 ! . Then :

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Correct answer: C. α ∈ N and β ∈ N

Given: α = ( 4 ! ) ! ( 4 ! ) 3 ! , β = ( 5 ! ) ! ( 5 ! ) 4 ! ⇒ α = ( 24 ) ! ( 4 ! ) 6 , β = ( 120 ) ! ( 5 ! ) 24 If 24 distinct objects are divided into 6 groups of 4 objects in each group, then the number of ways of formation of group is = 24 ! ( 4 ! ) 6 . 6 ! ∈ N Similarly, let 120 distinct objects are divided into 24 groups of 5 objects in each group. So, the number of ways of formation of groups = ( 120 ) ! ( 5 ! ) 24 · 24 ! ∈ N

Let α = ( 4 ! ) ! ( 4 ! ) 3 ! and β = ( 5 ! ) ! ( 5 ! ) 4 ! . Then :

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