Combinations

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

Correct result:

n =  11

Solution:

C2(11)=(112)=11!2!(112)!=111021=55 C4(11)=(114)=11!4!(114)!=1110984321=330 6 (n2)=(n4) 6 n 3/((n2)(n3)(n4)!)=1/(4 3 2)/(n4)! 4323=(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=b±D2a=5±2892 n1,2=5±172 n1,2=2.5±8.5 n1=11 n2=6   Factored form of the equation:  (n11)(n+6)=0 n>0 n=n1=11 C1=(n2)=55 C2=(n4)=330

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