Chapter 13.9, Problem 30PPS

(a)

To determine

To calculate: To complete the table

Answer to Problem 30PPS

Completed table is as follows:

Number of Points	Number of Segments
4	6
5	10
6	15
7	21

Explanation of Solution

Given information: In the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

  

Calculation:

Given that in the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

When number of points $=5$

Number of line segments $=10$

When number of points $=6$

Number of line segments $=15$

Completed table is as follows:

Number of Points	Number of Segments
4	6
5	$10$
6	$15$
7	21

Conclusion:

Hence, table is completed

(b)

To determine

To calculate: To explain how one can find the number of segments mathematically

Answer to Problem 30PPS

Mathematical formula to derive the number of segments is $\frac{n\left(n-1\right)}{2}$

Explanation of Solution

Given information: In the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

  

Calculation:

Given that in the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

Completed table is as follows:

Number of Points	Number of Segments
4	6
5	8
6	10
7	12

Mathematical formula is

  $\begin{array}{l}C\left(n,n-2\right)=\frac{n!}{\left(n-2\right)!\left(n-n+2\right)!}\\ C\left(n,n-2\right)=\frac{n!}{\left(n-2\right)!\left(2\right)!}\\ C\left(n,n-2\right)=\frac{n\times \left(n-1\right)\times \left(n-2\right)!}{\left(n-2\right)!\left(2\right)!}\\ C\left(n,n-2\right)=\frac{n\left(n-1\right)}{2}\end{array}$

Conclusion:

Hence, mathematical formula is $\frac{n\left(n-1\right)}{2}$

(c)

To determine

To calculate: To find $C\left(4,2\right),C\left(5,3\right),C\left(6,4\right),C\left(7,5\right)$ and compare the results to the number of segments for $4,5,6,7$ points

Answer to Problem 30PPS

  $C\left(4,2\right)=6$

  $C\left(5,3\right)=10$

  $C\left(7,5\right)=21$

  $C\left(6,4\right)=15$

Explanation of Solution

Given information: In the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

  

Formula Used:

  $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

  $C\left(4,2\right)$ is calculated as follows:

  $\begin{array}{l}C\left(4,2\right)=\frac{4!}{2!\left(4-2\right)!}\\ C\left(4,2\right)=\frac{4\times 3\times 2!}{2!2!}\\ C\left(4,2\right)=\frac{4\times 3}{2!}\\ C\left(4,2\right)=\frac{12}{2}\\ C\left(4,2\right)=6\end{array}$

This is same as number of segments for $4$ points

  $C\left(5,3\right)$ is calculated as follows:

  $\begin{array}{l}C\left(5,3\right)=\frac{5!}{3!\left(5-3\right)!}\\ C\left(5,3\right)=\frac{5\times 4\times 3!}{3!2!}\\ C\left(5,3\right)=\frac{5\times 4}{2!}\\ C\left(5,3\right)=\frac{20}{2}\\ C\left(5,3\right)=10\end{array}$

This is same as number of segments for $5$ points

  $C\left(6,4\right)$ is calculated as follows:

  $\begin{array}{l}C\left(6,4\right)=\frac{6!}{4!\left(6-4\right)!}\\ C\left(6,4\right)=\frac{6\times 5\times 4!}{4!2!}\\ C\left(6,4\right)=\frac{6\times 5}{2!}\\ C\left(6,4\right)=\frac{30}{2}\\ C\left(6,4\right)=15\end{array}$

This is same as number of segments for $6$ points

  $C\left(7,5\right)$ is calculated as follows:

  $\begin{array}{l}C\left(7,5\right)=\frac{7!}{5!\left(7-5\right)!}\\ C\left(7,5\right)=\frac{7\times 6\times 5!}{5!2!}\\ C\left(7,5\right)=\frac{7\times 6}{2!}\\ C\left(7,5\right)=\frac{42}{2}\\ C\left(7,5\right)=21\end{array}$

This is same as number of segments for $7$ points

Conclusion:

Hence, $C\left(4,2\right)=6$ , $C\left(5,3\right)=10$ , $C\left(7,5\right)=21$ and $C\left(6,4\right)=15$

(d)

To determine

To calculate: To predict the number of segments for $10$ , $12$ and $n$ points

Answer to Problem 30PPS

Number of segment for $10$ points is $45$

Number of segments for $12$ points is $66$

Number of segments for $n$ points is $\frac{n\left(n-1\right)}{2}$

Explanation of Solution

Given information: In the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

  

Calculation:

Given that in the figure, each of the $4$ non-collinear points are connected to every other points exactly once. There are two of six line segments connecting the points

Completed table is as follows:

Number of Points	Number of Segments
4	6
5	8
6	10
7	12

Number of segments for $n$ points is calculated as

  $\begin{array}{l}C\left(n,n-2\right)=\frac{n!}{\left(n-2\right)!\left(n-n+2\right)!}\\ C\left(n,n-2\right)=\frac{n!}{\left(n-2\right)!\left(2\right)!}\\ C\left(n,n-2\right)=\frac{n\times \left(n-1\right)\times \left(n-2\right)!}{\left(n-2\right)!\left(2\right)!}\\ C\left(n,n-2\right)=\frac{n\left(n-1\right)}{2}\end{array}$

Number of segments for $10$ point is calculated as

  $\begin{array}{l}C\left(10,8\right)=\frac{10!}{8!\left(10-8\right)!}\\ C\left(10,8\right)=\frac{10\times 9\times 8!}{8!2!}\\ C\left(10,8\right)=\frac{10\times 9}{2}\\ C\left(10,8\right)=\frac{90}{2}\\ C\left(10,8\right)=45\end{array}$

Number of segments for $12$ point is calculated as

  $\begin{array}{l}C\left(12,10\right)=\frac{12!}{10!\left(12-10\right)!}\\ C\left(12,10\right)=\frac{12\times 11\times 10!}{10!2!}\\ C\left(12,10\right)=\frac{12\times 11}{2}\\ C\left(12,10\right)=\frac{132}{2}\\ C\left(12,10\right)=66\end{array}$

Conclusion:

Hence, number of segment for $10$ points is $45$

Number of segments for $12$ points is $66$

Number of segments for $n$ points is $\frac{n\left(n-1\right)}{2}$