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:

6 V2(n)= V3(n) 6 n(n1) = n(n1)(n2) 6 = n2 n=6+2=8 V2=6 n (n1)=6 8 (81)=336 V3=n (n1) (n2)=8 (81) (82)=336

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