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

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 $x^{24}$)
