In how many ways can two dozen identical robots be assigned to four assembly lines with
(a) at least three robots assigned to each line?
(b) at least three, but no more than nine assigned to each line ?
Two dozen = 24,
Now we use generating functions so for (a) we would have $(x^3+x^4+.........)^4$ but we can take $x^{12}$ as a common factor so we have $(x^3+x^4+.........)^4 = x^{12}(1+x + x^2 + ........)^4$ and we already know that $\frac{1}{1-x} = (1+x+x^2+......)$ and so $\frac{1}{(1-x)^4} = (1+x+x^2+.......)^4$
and so we have $x^{12}(1+x + x^2 + ........)^4 = x^{12}(1-x)^{-4}$. Now want to find the coefficient of $x^{12}$ in $(1-x)^{-4}$ which is ${-4 \choose 12} = (-1)^{12} {4 + 12 -1 \choose 12} = {15 \choose 12}$
Now for (b) I get stuck.
First we have $(x^3 + x^4 + ........ +x^9)^4$, now again we take $x^{12}$ as a common factor and so we have $x^{12}(1 + x + x^2 + .... + x^6)^4$ , and now I want to find the coefficient of $x^{12}$ in $(1 + x + x^2 + .... + x^6)^4$, But how do we do this ?
Hint: We have, by the usual formula for the sum of a finite geometric progression, that if $x\ne 0$ then $$1+x+x^2+\cdots+x^6=\frac{1-x^7}{1-x}.$$ It follows that $$(1+x+x^2+\cdots+x^6)^4=(1-x^7)^4(1-x)^{-4}.$$ Note that we only need two easy terms of the expansion of $(1-x^7)^4$.