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There are $n$ songs segregated into $3$ play lists. Assume that each play list has at least one song.The number of ways of choosing three songs consisting of one song from each play list is:

  1.  $>\frac{n^3}{27}$ for all $n$
  2. $\leq \frac{n^3}{27}$ for all $n$
  3. $\left( \begin{array}{c} n \\ 3 \end{array} \right)$ for all $n$
  4. $n^3$ for all $n$
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Assume there are $a,b$ and $c$ numbers of songs in the first, second and third playlist respectively..

Therefore $a+b+c=n$. The number we can choose one song from each playlist is $abc$.

We have to find the maximum value of $abc$ corresponding to the condition $a+b+c=n$.

Let $L=abc-\lambda(a+b+c)$

Therefore $\frac{\partial L}{\partial a}=bc-\lambda, \frac{\partial L}{\partial b}=ac-\lambda, \frac{\partial L}{\partial c}=ab-\lambda$

For $\frac{\partial L}{\partial a}=\frac{\partial L}{\partial b}=\frac{\partial L}{\partial c}=0$ we get $ab=bc=cd=\lambda$ and $a=b=c$

Therefore $a^2b^2c^2=\lambda^3 \Rightarrow a^6=\lambda^3\Rightarrow a=\sqrt{\lambda}=b=c$

By the given condition $a+b+c=n\Rightarrow 3\sqrt{\lambda}=n\Rightarrow \sqrt{\lambda}=\frac{n}{3}$

Therefore the maximum value of $abc$ is $({\sqrt{\lambda}})^3=\frac{n^3}{27}$

Therefore the number of ways of choosing three songs consisting of one song from each play list is

  1. $\leq\frac{n^3}{27}$ for all $n$

 

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