2nd class variations

From how many elements can 5112 second-class variations be formed?

Final Answer:

n =  72

Step-by-step explanation:

V2(n)=5112 n (n1)=5112  n (n1)=5112 n2n5112=0  a=1;b=1;c=5112 D=b24ac=1241(5112)=20449 D>0  n1,2=2ab±D=21±20449 n1,2=21±143 n1,2=0.5±71.5 n1=72 n2=71  n>0 n=n1=72

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