Combinations

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

Final Answer:

n =  45

Step-by-step explanation:

 (2n)=   2} n(n1) = 990  n>0  n (n1)=2 990  n (n1)=2 990 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  n=n1=45

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