Question

Find the number of ways in which the number $30$ can be partitioned into three parts, each part being a natural number.

• A. $75$
• B. $61$
• C. $69$
• D. $70$

Answer 1

The correct option is B $61$

Let the three numbers be $x,y,z$ such that $x>y>z$

Let $x=y=z$ where $x-y=k>0$ and $y-z=j$ where $j>0$

$\Rightarrow x+y+z=3z+2j+k$ where $z<28$

$\therefore$ Number of solutions $=$ Coefficient of $x^{30}$ in $(x^3+x^6+...+x^{27})(x^2+x^4+...)(x+x^2+...)$

$=$ Coefficient of $x^{30}$ in $x^6(1-x^3{)}^{-1}(1-x^2{)}^{-1}(1-x{)}^{-1}$

$=61$