Find the number of ways in which the number 30 can be partitioned into three unequal parts.(Please use multinomial theorem) OfLet $a,b,c$ be the parts such that $a<b<c$
Now, let $a−b=x,     b−c=y,$   implies     $x,y>0$
$⇒a+b+c=30\\
⇒(b+x)+b+c=30\\
⇒x+2(c+y)+c=30\\
⇒x+2y+3c=30$      ,
$c≤27$
Sum =30,Co-eff =$1,2,3$
$(x^1+x^2+.......)×(x^2+x^4+.......)×(x^3+x^6+.......+x^{27})$
Co-efficient of $x^{30}$ in the above product will give us the required answer.=$x×x^2×x^3×(1−x)^{−1}×(1−x^2)^{−1}×(1−x^3)^{−1}\\
 =x^6\times (1−x)^{−1}\times(1−x^2)^{−1}\times(1−x^3)^{−1}$
Please help me to understand the last ,how to find the coefficient of p^24 in those 3 brackes
 A: You are pretty much on the right track. The last part is pretty tedious, so it might be better that you just use Wolfram Alpha or possibly some other program. I think you defined your variables a little wrong. For example, if $a<b$, then $a-b<0$. However, you defined $x=a-b$, but you are saying that $x>0$. It looks like you're fine if you instead force $c<b<a$.
As you found before, we want to find the coefficient of $x^{24}$ in
$$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Our goal will be to convert this into something of the form
$$\frac{P(x)}{(1-x^a)^b}$$
We can then use the fact that
$$\frac{1}{(1-x^a)^b}=\sum_{n=0}^\infty \binom{n+b-1}{b-1}x^{na}$$
to find the coefficient of $x^{24}$ in the resulting expansion. The most tedious part of this process is determining $P(x)$. We will start with some manipulations of
$$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Note that $1-x, 1-x^2,$ and $1-x^3$ are all factors of $1-x^6$ (I think this would be called the LCM). Using this, we get
$$=\frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)}{(1-x)(1+x+x^2+x^3+x^4+x^5)(1-x^2)(1+x^2+x^4)(1-x^3)(1+x^3)}$$
$$=\frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)}{(1-x^6)^3}$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
Well that was simple. However, the hard part is yet to come. We need to find the coefficient of $x^{24}$ in that expansion. Denote the product of the leftmost $3$ polynomials (the ones not in the summation) as $P(x)$. We will be finding the coefficient $x^{24}$ in the expansion of
$$P(x)\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
Since $\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$ only has terms of powers $0\mod 6$, we only care about the terms with power $0\mod 6$ in the expansion of $P(x)$. Although this doesn't help us too much, it does reduce some of the calculations. Expanding $P(x)$ while only caring about the final terms whose powers are $0\mod 6$ gives
$$P(x)$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^3+x^4+x^5+x^7)$$
$$\sim 1+x^6+x^6+x^6+x^6+x^{12}$$
$$=1+4x^6+x^{12}$$
So we only need to find the coefficient of $x^{24}$ in
$$(1+4x^6+x^{12})\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
$$\binom{4+2}{2}+4\cdot \binom{3+2}{2}+\binom{2+2}{2}$$
$$=15+40+6$$
$$=61$$
Indeed this matches what Wolfram Alpha gives (just click more terms on the taylor series expansion until you get to $x^{24}$)
