# Variations 4/2

Determine the number of items when the count of variations of fourth class without repeating is 600 times larger than the count of variations of second class without repetition.

Correct result:

n =  27

#### Solution:

$V_4(n)= 600 \cdot V_2(n) \ \\ n(n-1)(n-2)(n-3) = 600 \cdot n(n-1) \ \\ (n-2)(n-3) = 600 \ \\ n^2 -5n +6 - 600 = 0 \ \\ n_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 5 \pm \sqrt{ 2401 } }{ 2 } \ \\ n_{1,2} = \dfrac{ 5 \pm 49 }{ 2 } \ \\ n_{1,2} = 2.5 \pm 24.5 \ \\ n_{1} = 27 \ \\ n_{2} = -22 \ \\ n>0 \ \\ n = 27 \ \\$

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