In the market have 3 kinds of chocolates.

How many ways can we buy 14 chocolates?


n =  120


n=C143=C14(16)=(1614)=16!14!(1614)!=161521=120n = C'^{ 3}_{ 14} = C_{{ 14}}(16) = \dbinom{ 16}{ 14} = \dfrac{ 16! }{ 14!(16-14)!} = \dfrac{ 16 \cdot 15 } { 2 \cdot 1 } = 120

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