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  4. If a, b, c are three natural numbers in AP and a+b+c=21 then the possible number of values of the ordered triplet (a, b, c) is

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Q. If $a, b, c$ are three natural numbers in $AP$ and $a+b+c=21$ then the possible number of values of the ordered triplet $(a, b, c)$ is

BITSATBITSAT 2020

Solution:

Let $a=b-d$ and $c=b+d$, then $a+b+c=21$
$\Rightarrow b=7$
So, the equation is $a+c=14$
$\therefore $ No. of solution $=$ coeff. of $x^{14}$ in $\left(x+x^{2}+\ldots .\right)$
$={ }^{13} C_{12}=13$