Combination element calculation

How many elements is it possible to form twice as many second-class combinations as a fourth-class combination?

Final Answer:

n =  5

Step-by-step explanation:

(2n) = 2 (4n) n /2 1 /((n2)(n3)(n4)!) = 4/(4 3 2)/(n4)! 3 2=(n2)(n3)  3 2=(n2)(n3) n2+5n=0 n25n=0  a=1;b=5;c=0 D=b24ac=52410=25 D>0  n1,2=2ab±D=25±25 n1,2=25±5 n1,2=2.5±2.5 n1=5 n2=0  n>0 n=n1=5

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combinatoricsalgebraGrade of the word problem

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