Combinations

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

Correct answer:

n =  11

Step-by-step explanation:

C2​(11)=(211​)=2!(11−2)!11!​=2⋅111⋅10​=55 C4​(11)=(411​)=4!(11−4)!11!​=4⋅3⋅2⋅111⋅10⋅9⋅8​=330 6⋅ (2n​) = (4n​) 6⋅ n /2 3 /((n−2)(n−3)(n−4)!) = 1/(4⋅ 3⋅ 2)/(n−4)! 4⋅ 3⋅ 2⋅ 3=(n−2)(n−3)  4⋅ 3⋅ 2⋅ 3=(n−2)(n−3) −n2+5n+66=0 n2−5n−66=0  a=1;b=−5;c=−66 D=b2−4ac=52−4⋅1⋅(−66)=289 D>0  n1,2​=2a−b±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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