Question

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

• A. $61$
• B. $64$
• C. $65$
• D. $55$

Answer 1

Answer: (A)

$61$

$30$ must be divided into $3$ unequal parts.

Let $a,b,c$ be the parts such that $a<b<c$

Now, let $a-b=x,$ $b-c=y,$ $x,y>0$

$\Rightarrow a+b+c=30$

$\Rightarrow (b+x)+b+c=30$

$\Rightarrow x+2(c+y)+c=30$

$\Rightarrow x+2y+3c=30$ $,$ $c\le 27$

Sum $=30,$ Co-eff $=1,2,3$

$(x^1+x^2+.......)\times (x^2+x^4+.......)\times (x^3+x^6+.......+x^{27})$

Co-efficient of $x^{30}$ in the above product will give us the required answer.

$=x\times x^2\times x^3\times {(1-x)}^{-1}\times {(1-x^2)}^{-1}\times {(1-x^3)}^{-1}$

$=x^6{(1-x)}^{-1}{(1-x^2)}^{-1}{(1-x^3)}^{-1}$

$=61$

Hence, the answer is $61.$