The number of ways of selecting two distinct numbers from the first $15$ natural numbers such that their sum is a multiple of $\mathrm{5,}$ is equal to

$x+y=5,\mathrm{}10,\mathrm{}15,\mathrm{}20,\mathrm{}25,\mathrm{}30$
Where $x\ne y,\mathrm{}\left(x,y\right)=\left(y,x\right)$
The number of solutions of $x+y=5$ is $\frac{4}{2!}$
The number of solutions of $x+y=10$ is $\frac{9-1}{2!}$
The number of solutions of $x+y=15$ is $\frac{14}{2!}$
The number of solutions of $x+y=20$ is $\frac{11-1}{2!}$
$\left(\because x=5,\mathrm{}6,\dots ,\mathrm{}15\right)$
The number of solutions of $x+y=25$ is $\frac{6}{2!}$
$\left(\because x=10,\mathrm{}11,\mathrm{}\dots ,\mathrm{}15\right)$
The number of solutions of $x+y=30$ is $0$
$\left(\because x=y=15\right)$
So, the required number of solutions
$=2+4+7+5+3+0=21$