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3

$begingroup$

Suppose we are to find the number of triangles that exist from the given figure

I found one solution that says we let $A$ equal the number of internal lines from the top vertex, $B$ equal the number of internal lines parallel to the base, using the formula below to find the number of $N$ triangles
$$N=frac(A+1)(A+2)(B+1)2$$

With $A$ equal to 2 and $B$ equal to 3, we get $N=24$ triangles.

But can somebody explain why this formula works? How exactly do I derive this?

combinatorics algebra-precalculus geometry

share|cite|improve this question

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Suppose we are to find the number of triangles that exist from the given figure

I found one solution that says we let $A$ equal the number of internal lines from the top vertex, $B$ equal the number of internal lines parallel to the base, using the formula below to find the number of $N$ triangles
$$N=frac(A+1)(A+2)(B+1)2$$

With $A$ equal to 2 and $B$ equal to 3, we get $N=24$ triangles.

But can somebody explain why this formula works? How exactly do I derive this?

combinatorics algebra-precalculus geometry

share|cite|improve this question

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

$endgroup$

add a comment | 

$begingroup$

Suppose we are to find the number of triangles that exist from the given figure

I found one solution that says we let $A$ equal the number of internal lines from the top vertex, $B$ equal the number of internal lines parallel to the base, using the formula below to find the number of $N$ triangles
$$N=frac(A+1)(A+2)(B+1)2$$

With $A$ equal to 2 and $B$ equal to 3, we get $N=24$ triangles.

But can somebody explain why this formula works? How exactly do I derive this?

combinatorics algebra-precalculus geometry

share|cite|improve this question

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

share|cite|improve this question

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

edited Jun 10 at 9:37

user21820

41.1k55 gold badges4646 silver badges167167 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

asked Jun 10 at 0:44

Lex_iLex_i

16077 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

A triangle inside the figure has one of its sides being one of the $B+1$ horizontal lines, and the other two sides are two of the $A+2$ slant lines. So the total number of ways of forming a triangle is $displaystyle A+2 choose 2 times B+1 choose 1$.

share|cite|improve this answer

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you multiply the matrices together for me to show it getting 24? Because when I do it I keep getting 18
  $endgroup$
  – Lex_i
  Jun 10 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  They are binomial coefficients, not matrices. $n choose r$ is the number of ways of choosing $r$ objects from $n$ objects and the formula is $nchoose r=fracn!r!(n-r)!$. In particular, $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1=B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $n choose 1=fracn!(n-1)!1!=fracntimes (n-1)!(n-1)!times 1=n$ and $n choose 2=fracn!(n-2)!2!=fracn(n-1)times (n-2)!(n-2)!times 2=fracn(n-1)2$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1$ is just $B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:51
  

  
  
  
  

add a comment | 

0

$begingroup$

If we let $x$ to be the number of lines from the top vertex and $y$ to be the number of lines parallel to the base including the base, then we have $x = A+2$ and $y = B+1$. We keep this.

Now, think about how can we construct a triangle using these lines. First, we need a base, which we can choose from $y$ lines. Then, we need two other lines to construct a triangle but these two lines should intersect at one point (which is top vertex in this case). Therefore, we can actually choose $2$ lines from $x$ lines since all of these $x$ lines intersect at top vertex. So we can choose these $2$ lines with $binomx2 = fracx(x-1)2$. So, in total, we can construct $fracx(x-1)y2$ different triangles. Now, use $x = A+2$, $y = B+1$.

share|cite|improve this answer

answered Jun 10 at 0:54

ArsenBerkArsenBerk

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$endgroup$

add a comment | 

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6

$begingroup$

A triangle inside the figure has one of its sides being one of the $B+1$ horizontal lines, and the other two sides are two of the $A+2$ slant lines. So the total number of ways of forming a triangle is $displaystyle A+2 choose 2 times B+1 choose 1$.

share|cite|improve this answer

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you multiply the matrices together for me to show it getting 24? Because when I do it I keep getting 18
  $endgroup$
  – Lex_i
  Jun 10 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  They are binomial coefficients, not matrices. $n choose r$ is the number of ways of choosing $r$ objects from $n$ objects and the formula is $nchoose r=fracn!r!(n-r)!$. In particular, $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1=B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $n choose 1=fracn!(n-1)!1!=fracntimes (n-1)!(n-1)!times 1=n$ and $n choose 2=fracn!(n-2)!2!=fracn(n-1)times (n-2)!(n-2)!times 2=fracn(n-1)2$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1$ is just $B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:51
  

  
  
  
  

add a comment | 

6

$begingroup$

A triangle inside the figure has one of its sides being one of the $B+1$ horizontal lines, and the other two sides are two of the $A+2$ slant lines. So the total number of ways of forming a triangle is $displaystyle A+2 choose 2 times B+1 choose 1$.

share|cite|improve this answer

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you multiply the matrices together for me to show it getting 24? Because when I do it I keep getting 18
  $endgroup$
  – Lex_i
  Jun 10 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  They are binomial coefficients, not matrices. $n choose r$ is the number of ways of choosing $r$ objects from $n$ objects and the formula is $nchoose r=fracn!r!(n-r)!$. In particular, $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1=B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $n choose 1=fracn!(n-1)!1!=fracntimes (n-1)!(n-1)!times 1=n$ and $n choose 2=fracn!(n-2)!2!=fracn(n-1)times (n-2)!(n-2)!times 2=fracn(n-1)2$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $A+2 choose 2=frac(A+2)(A+1)2$ and $B+1 choose 1$ is just $B+1$.
  $endgroup$
  – CY Aries
  Jun 10 at 2:51
  

  
  
  
  

add a comment | 

$begingroup$

A triangle inside the figure has one of its sides being one of the $B+1$ horizontal lines, and the other two sides are two of the $A+2$ slant lines. So the total number of ways of forming a triangle is $displaystyle A+2 choose 2 times B+1 choose 1$.

share|cite|improve this answer

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

answered Jun 10 at 0:53

CY AriesCY Aries

21.3k11 gold badge2020 silver badges4949 bronze badges

0

$begingroup$

If we let $x$ to be the number of lines from the top vertex and $y$ to be the number of lines parallel to the base including the base, then we have $x = A+2$ and $y = B+1$. We keep this.

Now, think about how can we construct a triangle using these lines. First, we need a base, which we can choose from $y$ lines. Then, we need two other lines to construct a triangle but these two lines should intersect at one point (which is top vertex in this case). Therefore, we can actually choose $2$ lines from $x$ lines since all of these $x$ lines intersect at top vertex. So we can choose these $2$ lines with $binomx2 = fracx(x-1)2$. So, in total, we can construct $fracx(x-1)y2$ different triangles. Now, use $x = A+2$, $y = B+1$.

share|cite|improve this answer

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

If we let $x$ to be the number of lines from the top vertex and $y$ to be the number of lines parallel to the base including the base, then we have $x = A+2$ and $y = B+1$. We keep this.

Now, think about how can we construct a triangle using these lines. First, we need a base, which we can choose from $y$ lines. Then, we need two other lines to construct a triangle but these two lines should intersect at one point (which is top vertex in this case). Therefore, we can actually choose $2$ lines from $x$ lines since all of these $x$ lines intersect at top vertex. So we can choose these $2$ lines with $binomx2 = fracx(x-1)2$. So, in total, we can construct $fracx(x-1)y2$ different triangles. Now, use $x = A+2$, $y = B+1$.

share|cite|improve this answer

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges

$endgroup$

add a comment | 

$begingroup$

If we let $x$ to be the number of lines from the top vertex and $y$ to be the number of lines parallel to the base including the base, then we have $x = A+2$ and $y = B+1$. We keep this.

Now, think about how can we construct a triangle using these lines. First, we need a base, which we can choose from $y$ lines. Then, we need two other lines to construct a triangle but these two lines should intersect at one point (which is top vertex in this case). Therefore, we can actually choose $2$ lines from $x$ lines since all of these $x$ lines intersect at top vertex. So we can choose these $2$ lines with $binomx2 = fracx(x-1)2$. So, in total, we can construct $fracx(x-1)y2$ different triangles. Now, use $x = A+2$, $y = B+1$.

share|cite|improve this answer

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges

answered Jun 10 at 0:54

ArsenBerkArsenBerk

9,65033 gold badges1414 silver badges3939 bronze badges