Variations 4/2

Determine the number of items when the count of variations of the fourth class without repeating is 5112 times larger than the count of variations of the second class 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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