From how many elements we can create 990 combinations 2nd class without repeating?

Correct answer:

n =  45

Step-by-step explanation:

(2n)=2n(n1)=990;n>0 n2n1980=0  a=1;b=1;c=1980 D=b24ac=1241(1980)=7921 D>0  n1,2=2ab±D=21±7921 n1,2=21±89 n1,2=0.5±44.5 n1=45 n2=44   Factored form of the equation:  (n45)(n+44)=0   n>0n=45

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