On the menu are 12 kinds of meal. How many ways can we choose four different meals into the daily menu?

Result

n =  495

#### Solution:

$C_{{ 4}}(12)=\dbinom{ 12}{ 4}=\dfrac{ 12! }{ 4!(12-4)!}=\dfrac{ 12 \cdot 11 \cdot 10 \cdot 9 } { 4 \cdot 3 \cdot 2 \cdot 1 }=495 \ \\ \ \\ n={ { 12 } \choose 4 }=495$

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