If the number of elements increase by 3, it increases the number of combinations of the second class of these elements 5 times. How many are the elements?

Correct result:

n =  3


C2(6)=(62)=6!2!(62)!=6521=15 5 (n2)=(n+32) 5 n 5n(n1)=(n+3)(n+2)  5 n(n1)=(n+3) (n+2) 4n210n6=0  a=4;b=10;c=6 D=b24ac=10244(6)=196 D>0  n1,2=b±D2a=10±1968 n1,2=10±148 n1,2=1.25±1.75 n1=3 n2=0.5   Factored form of the equation:  4(n3)(n+0.5)=0 n>0 n=n1=3 C1=(n2)=3 C2(3)=(32)=3!2!(32)!=31=3  C2=(n+32)=(3+32)=15 n=3

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