Variations Class Comparison

From how many elements can we create six times as many variations of the second class without repetition as variations of the third class without repetition?

Final Answer:

n =  8

Step-by-step explanation:

$$\begin{gathered} 6\cdot V_2(n)=V_3(n) \\ 6\cdot n(n-1)=n(n-1)(n-2) \\ 6=n-2 \\ n=6+2=8 \\ {V}_2=6\cdot n\cdot (n-1)=6\cdot 8\cdot (8-1)=336 \\ {V}_3=n\cdot (n-1)\cdot (n-2)=8\cdot (8-1)\cdot (8-2)=336 \end{gathered}$$

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