How many different samples of size n = 3 can be selected from a population with the following sizes?
1.) N = 4
2.) N = 8
3.) N = 20
4.) N = 50
The number of possible samples which can be selected without replacement is
(Nn)=N!n!⋅(N−n)!.\begin{pmatrix} N \\ n \end{pmatrix}=\cfrac{N! } {n! \cdot(N-n)! }.(Nn)=n!⋅(N−n)!N!.
1.) (43)=4!3!⋅1!=4.1.) \ \begin{pmatrix} 4 \\ 3 \end{pmatrix}=\cfrac{4! } {3! \cdot1! }=4.1.) (43)=3!⋅1!4!=4.
2.) (83)=8!3!⋅5!==6⋅7⋅82⋅3=56.2.) \ \begin{pmatrix} 8\\ 3 \end{pmatrix}=\cfrac{8! } {3! \cdot5! }=\\ =\cfrac{6\cdot7\cdot8}{2\cdot3}=56.2.) (83)=3!⋅5!8!==2⋅36⋅7⋅8=56.
3.) (203)=20!3!⋅17!==18⋅19⋅202⋅3=1140.3.) \ \begin{pmatrix} 20\\ 3 \end{pmatrix}=\cfrac{20! } {3! \cdot17! }=\\ =\cfrac{18\cdot19\cdot20}{2\cdot3}=1140.3.) (203)=3!⋅17!20!==2⋅318⋅19⋅20=1140.
4.) (503)=50!3!⋅47!==48⋅49⋅502⋅3=19600.4.) \ \begin{pmatrix} 50\\ 3 \end{pmatrix}=\cfrac{50! } {3! \cdot47! }=\\ =\cfrac{48\cdot49\cdot50}{2\cdot3}=19600.4.) (503)=3!⋅47!50!==2⋅348⋅49⋅50=19600.
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