Mixing
Count number of triangles given a set of points
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Gievn, $$A = le 2$$
Find
(a) Number of straight lines which pass through at least 2 points in $A$;
(b) Number of triangles whose vertices are points in $A$;
(a) I have exhaustively counted it's answer and it turns out to be 40. But it there any general approach to solve it?
(b) Same exhaustive approach: First choosing one point on a line then choosing other two points from remaining points. Second choosing 2 points on a line and one point that is not on the line. Sum these two cases for every line.
Is there any general approach for this as well?
combinatorics
asked Jul 14 at 16:37
naman jain
185
add a comment |Â
up vote
1
down vote
favorite
Gievn, $$A = le 2$$
Find
(a) Number of straight lines which pass through at least 2 points in $A$;
(b) Number of triangles whose vertices are points in $A$;
(a) I have exhaustively counted it's answer and it turns out to be 40. But it there any general approach to solve it?
(b) Same exhaustive approach: First choosing one point on a line then choosing other two points from remaining points. Second choosing 2 points on a line and one point that is not on the line. Sum these two cases for every line.
Is there any general approach for this as well?
combinatorics
asked Jul 14 at 16:37
naman jain
185
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Gievn, $$A = le 2$$
Find
(a) Number of straight lines which pass through at least 2 points in $A$;
(b) Number of triangles whose vertices are points in $A$;
(a) I have exhaustively counted it's answer and it turns out to be 40. But it there any general approach to solve it?
(b) Same exhaustive approach: First choosing one point on a line then choosing other two points from remaining points. Second choosing 2 points on a line and one point that is not on the line. Sum these two cases for every line.
Is there any general approach for this as well?
combinatorics
asked Jul 14 at 16:37
naman jain
185
Gievn, $$A = le 2$$
Find
(a) Number of straight lines which pass through at least 2 points in $A$;
(b) Number of triangles whose vertices are points in $A$;
(a) I have exhaustively counted it's answer and it turns out to be 40. But it there any general approach to solve it?
(b) Same exhaustive approach: First choosing one point on a line then choosing other two points from remaining points. Second choosing 2 points on a line and one point that is not on the line. Sum these two cases for every line.
Is there any general approach for this as well?
combinatorics
asked Jul 14 at 16:37
naman jain
185
edited Jul 14 at 20:22
user401938
asked Jul 14 at 16:37
naman jain
185
asked Jul 14 at 16:37
naman jain
185
asked Jul 14 at 16:37
naman jain
185
185
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
Drawing a figure we see $10$ lines with $3$ points and $2$ lines with $5$ points. All other lines in the plane contain at most $2$ of the given $13$ points.
(a) "Formally" there are $13choose2=78$ lines connecting two points in $A$. Lines containing $3$ points of $A$ are thereby counted $3choose2=3$ times, and lines containing $5$ points of $A$ are counted $5choose2=10$ times. It follows that the number of different lines is given by
$$N_rm lines=78-10(3-1)-2(10-1)=40 .$$
(b) There are $13choose3=286$ triples of points from $A$. Triples lying on a line do not give rise to a nondegenerate triangle. It follows that the number of nondegenerate triangles in this figure is given by
$$N_rm triangles=286-103choose3-25choose3=256 .$$
answered Jul 14 at 19:13
Christian Blatter
164k7108306
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Drawing a figure we see $10$ lines with $3$ points and $2$ lines with $5$ points. All other lines in the plane contain at most $2$ of the given $13$ points.
(a) "Formally" there are $13choose2=78$ lines connecting two points in $A$. Lines containing $3$ points of $A$ are thereby counted $3choose2=3$ times, and lines containing $5$ points of $A$ are counted $5choose2=10$ times. It follows that the number of different lines is given by
$$N_rm lines=78-10(3-1)-2(10-1)=40 .$$
(b) There are $13choose3=286$ triples of points from $A$. Triples lying on a line do not give rise to a nondegenerate triangle. It follows that the number of nondegenerate triangles in this figure is given by
$$N_rm triangles=286-103choose3-25choose3=256 .$$
answered Jul 14 at 19:13
Christian Blatter
164k7108306
add a comment |Â
up vote
2
down vote
Drawing a figure we see $10$ lines with $3$ points and $2$ lines with $5$ points. All other lines in the plane contain at most $2$ of the given $13$ points.
(a) "Formally" there are $13choose2=78$ lines connecting two points in $A$. Lines containing $3$ points of $A$ are thereby counted $3choose2=3$ times, and lines containing $5$ points of $A$ are counted $5choose2=10$ times. It follows that the number of different lines is given by
$$N_rm lines=78-10(3-1)-2(10-1)=40 .$$
(b) There are $13choose3=286$ triples of points from $A$. Triples lying on a line do not give rise to a nondegenerate triangle. It follows that the number of nondegenerate triangles in this figure is given by
$$N_rm triangles=286-103choose3-25choose3=256 .$$
answered Jul 14 at 19:13
Christian Blatter
164k7108306
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Drawing a figure we see $10$ lines with $3$ points and $2$ lines with $5$ points. All other lines in the plane contain at most $2$ of the given $13$ points.
(a) "Formally" there are $13choose2=78$ lines connecting two points in $A$. Lines containing $3$ points of $A$ are thereby counted $3choose2=3$ times, and lines containing $5$ points of $A$ are counted $5choose2=10$ times. It follows that the number of different lines is given by
$$N_rm lines=78-10(3-1)-2(10-1)=40 .$$
(b) There are $13choose3=286$ triples of points from $A$. Triples lying on a line do not give rise to a nondegenerate triangle. It follows that the number of nondegenerate triangles in this figure is given by
$$N_rm triangles=286-103choose3-25choose3=256 .$$
answered Jul 14 at 19:13
Christian Blatter
164k7108306
Drawing a figure we see $10$ lines with $3$ points and $2$ lines with $5$ points. All other lines in the plane contain at most $2$ of the given $13$ points.
(a) "Formally" there are $13choose2=78$ lines connecting two points in $A$. Lines containing $3$ points of $A$ are thereby counted $3choose2=3$ times, and lines containing $5$ points of $A$ are counted $5choose2=10$ times. It follows that the number of different lines is given by
$$N_rm lines=78-10(3-1)-2(10-1)=40 .$$
(b) There are $13choose3=286$ triples of points from $A$. Triples lying on a line do not give rise to a nondegenerate triangle. It follows that the number of nondegenerate triangles in this figure is given by
$$N_rm triangles=286-103choose3-25choose3=256 .$$
answered Jul 14 at 19:13
Christian Blatter
164k7108306
answered Jul 14 at 19:13
Christian Blatter
164k7108306
answered Jul 14 at 19:13
Christian Blatter
164k7108306
answered Jul 14 at 19:13
Christian Blatter
164k7108306
164k7108306
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2851754%2fcount-number-of-triangles-given-a-set-of-points%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password