Variations 5437

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?

Correct 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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