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I was reading a course on Combinatorics where I came across following:

The number of ways in which $n$ identical things can be divided into $r$ groups so that no group contains less than $m$ items and more than $k$ (where $m<k$) is coefficient of $x^n$ in the expansion of $(x^m + x^{m+1} + … +x^k)^r$.

Is this correct? Book doesn't give proof. How can we make this formula intuitive?

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  • $\begingroup$ If you factor out $x^m$ you get the easy case where you have to divide $n-rm$ identical things into $r$ groups. $\endgroup$ – user53970 May 22 '15 at 17:38
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Write $$\underbrace{(x^m + x^{m+1} + … +x^k)(x^m + x^{m+1} + … +x^k)\cdots(x^m + x^{m+1} + … +x^k)}_{r\text{ times}}$$

In order to get a $x^n$ term you must choose a term from each parethesis in a such way that the sum of the exponents is $n$. This is the same as writing $n$ as the sum of $r$ numbers between $m$ and $k$.

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  • $\begingroup$ I still didnt get to know from where that $x^n$ term came, and in fact I didnt get from where that whole polynomial came. Does your answer explains that? I think I can do the expansion and get the coefficient of $x^n$ in it. But I dont understand from where all that $x^n$ thingy came. Pardon me for my poor understanding. $\endgroup$ – anir123 May 22 '15 at 20:29

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