Question

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

• A. $20$
• B. $36$
• C. $21$
• D. $16$

Answer 1

Answer: (C)

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