Variations

Determine the number of items when the count of variations of fourth class without repeating is 42 times larger than the count of variations of third class without repetition.

Result

n =  45

Solution:

V4(n)=42⋅V3(n) n(n−1)(n−2)(n−3)=42⋅n(n−1)(n−2) (n−3)=42 n=42+3=45V_4(n)= 42 \cdot V_3(n) \ \\ n(n-1)(n-2)(n-3) = 42 \cdot n(n-1)(n-2) \ \\ (n-3) = 42 \ \\ n = 42+3 = 45

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