Variations 4/2

Determine the number of elements such that the number of fourth-class variations without repetition is 5112 times greater than the number of second-class variations without repetition.

Final Answer:

n =  74

Step-by-step explanation:

$$\begin{gathered} V_4(n)=5112\cdot V_2(n) \\ n(n-1)(n-2)(n-3)=5112\cdot n(n-1) \\ (n-2)(n-3)=5112 \\ {n}^2-5n+6-5112=0 \\ {n}^2-5n-5106=0 \\ a=1;b=-5;c=-5106 \\ D=b^2-4ac=5^2-4\cdot 1\cdot (-5106)=20449 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{5\pm \sqrt{20449}}{2} \\ {n}_{1,2}=\frac{5\pm 143}{2} \\ {n}_{1,2}=2.5\pm 71.5 \\ {n}_1=74 \\ {n}_2=-69 \\ n>0 \\ n=n_1=74 \end{gathered}$$

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