BdMO Online Forum

We can take $1,2$ or $0$ points from the $5$ collinear points and the others from the rest $10$ points to make triangle and to make straight line we can take $1$ or $0$ points from the $5$ collinear points and the others from the rest $10$ points.

Prove it:If a graph has $e$ edges and $v$ vertices and $e≥v$,then the graph must contain a $cycle$.

Let $S={x: x=3k,x<21}$;$B={y: y=3k+1,y<21}$;$C={z: z=3k+2,z<21}$.
Now think how can we make $3$ element sets which satisfie the question.

We can take $1,2$ or $0$ points from the $5$ collinear points and the others from the rest $10$ points to make triangle.So,the number of total triangles that can be made by using the $15$ points is $\binom{5}{0}\binom{10}{3}+\binom{5}{1}\binom{10}{2}+\binom{5}{2}\binom{10}{1}=445$;Now to make straight line we can take $1$ or $0$ points from the $5$ collinear points and the others from the rest $10$ points.So,the number of total straight lines that can be made by using the $15$ points is $\binom{5}{0}\binom{10}{2}+\binom{5}{1}\binom{10}{1}=95+1=96$.

Let $S={x: x=3k,x<21}$;$B={y: y=3k+1,y<21}$;$C={z: z=3k+2,z<21}$.Now,we can make $3$ element subsets that satisfy the condition in the following ways:
(1)We can take $3$ numbers from $S$.
(2)We can take $1$ number from $S$,$1$ number from $B$,$1$ number from $C$.
(3)We can take $3$ numbers from $B$.
(4)We can take $3$ numbers from $C$.
So,the total numbers of subsets which satisfy the condition is $\binom{6}{3}+\binom{6}{1}\binom{7}{1}\binom{7}{1}+\binom{7}{3}+\binom{7}{3}=384$.
.