Combinations

How many elements can form six times more combinations fourth class than the combination of the second class?

Correct answer:

n =  11

Step-by-step explanation:

C2(11)=(211)=2!(112)!11!=211110=55 C4(11)=(411)=4!(114)!11!=4321111098=330 6 (2n) = (4n) 6 n /2 3 /((n2)(n3)(n4)!) = 1/(4 3 2)/(n4)! 4 3 2 3=(n2)(n3)  4 3 2 3=(n2)(n3) n2+5n+66=0 n25n66=0  a=1;b=5;c=66 D=b24ac=5241(66)=289 D>0  n1,2=2ab±D=25±289 n1,2=25±17 n1,2=2.5±8.5 n1=11 n2=6 n>0 n=n1=11 C1=(2n)=55 C2=(4n)=330

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