In how many ways can three songs be selected from $n$ songs segregated into $3$ playlists if each playlist has at least one song? There are $n$ songs segregated into $3$ playlists. Assume that each playlist has at least one song. The number of ways of choosing three songs consisting of one song from each playlist is:
Please help. I'm thinking about this problem from a long time.
Do we have to find the number of integer solutions?
 A: The question I am answer is this: 
there are $n$ songs. Now find the number of ways to choose $3$ songs in the following manner.
divide the $n$ songs into $3$ playlists, such that each playlist has at-least one song. then choose $1$ song from each playlist.
Ans:
choose 1st song. put it in playlist number 1. choose 2nd song. put it in playlist number 2. choose 3rd song. put it in playlist number 3.
the 3 songs chosen until now, are the final 3 songs you are asked to choose.
Notice each playlist right now, has exactly one song.
Now, segregate the remaining $n-3$ songs into the 3 playlists, any way you like(of course repetition is not allowed).
so the answer is $n\cdot (n-1)\cdot (n-2)\cdot \binom{n-3+(3-1)}{(3-1)}=\binom{n}{3}3!\cdot\binom{n-1}{2}$
Now, notice that each possibility has been counted 6 times.
so the final answer is $\binom{n}{3}\cdot\binom{n-1}{2}$
Note: 
this question was asked in a certain exam. A lot of people misunderstood what was being asked, including me. What was being asked is that, you are given $n$ songs. These $n$ songs have already been divided into $3$ playlists for you, such that each playlist has at-least one song. Now given these 3 playlists, you have to find the correct options.
A: Forget about the playlists. And just pike 3 songs combinations. So the result is: $$\binom{n}{3}$$
A: I know I am maybe answering a little too late, but the number of playlist does matter here.
We are segregating n songs into 3 playlists, with at least one song in each playlist. Assuming the perfect case (when n is a multiple of 3) we can best divide the songs into 3 playlists with each containing $\frac{n}{3}$ number of songs.
And thus, clearly, the number of ways we can choose 3 songs with one from each playlist is
$\frac{n}{3}*\frac{n}{3}*\frac{n}{3}$ = $\frac{n^3}{27}$
When n is not a multiple of 3 (or when the songs are not divided equally into 3 playlists), the number of ways we can choose the songs must be $\lt\frac{n^3}{27}$
Thus, our ultimate answer of the number of ways is $\le\frac{n^3}{27}$
