Question

The number of straight lines that can be drawn out of \[10\] points of which \[7\] are collinear is
\[\left( 1 \right){\text{ }}22\]
\[\left( 2 \right){\text{ 23}}\]
\[\left( 3 \right){\text{ 24}}\]
\[\left( 4 \right){\text{ 25}}\]

Answer 1

Hint: Collinear points are the points that lie on the same line. Whereas the points which do not lie on the same line are called non-collinear points. In the question it is given that $7$ points are collinear , so we can say that these $7$ points are together in the same line and the remaining 3 points are non-collinear points. These $7$ points together can give only 1 straight line as all the points are collinear.

Complete step-by-step solution:
We have to find how many straight lines can be possibly drawn out of given $10$ points from which $7$ points are collinear and the remaining $3$ points are non-collinear points.
We know that to make a line we need at least $2$ points. So, the number of ways to select $2$ points from $10$ points to make straight lines is \[{}^{10}{C_2}\]
As it is given that $7$ points are collinear which means that we can’t make lines for all the $7$ points. So, the number of ways to select $2$ points from these $7$ points which will not give any distinct line is \[{}^7{C_2}\]
So, we will subtract \[{}^7{C_2}\] from \[{}^{10}{C_2}\] . But $7$ collinear points are on a straight line. Therefore we have to add \[ + 1\] in \[{}^{10}{C_2}\] .
\[\therefore \] Total number of lines \[ = {\text{ }}{}^{10}{C_2} - {}^7{C_2} + 1\]
 We know that the number of combinations of n objects taken r at a time is determined by the formula \[{}^n{C_r} = \dfrac{{n!}}{{n!\left( {n - r} \right)!}}\] . So, by applying this formula we get
\[ \Rightarrow {\text{ }}\dfrac{{10!}}{{2!{\text{ }} \times {\text{ }}8!}} - \dfrac{{7!}}{{2!{\text{ }} \times {\text{ }}5!}} + 1\]
\[ \Rightarrow {\text{ }}\dfrac{{10{\text{ }} \times {\text{ }}9{\text{ }} \times {\text{ }}8!}}{{2{\text{ }} \times {\text{ }}1{\text{ }} \times {\text{ }}8!}} - \dfrac{{7{\text{ }} \times {\text{ }}6{\text{ }} \times {\text{ }}5!}}{{2{\text{ }} \times {\text{ }}1{\text{ }} \times {\text{ }}5!}} + 1\]
Now, \[8\] factorial and \[5\] factorial we will be cancelled out and we get
\[ \Rightarrow {\text{ }}\dfrac{{10{\text{ }} \times {\text{ }}9{\text{ }}}}{{2{\text{ }}}} - \dfrac{{7{\text{ }} \times {\text{ }}6{\text{ }}}}{{2{\text{ }}}} + 1\]
\[ \Rightarrow {\text{ }}\dfrac{{90{\text{ }}}}{{2{\text{ }}}} - \dfrac{{{\text{42 }}}}{{2{\text{ }}}} + 1\]
Divide the numbers by $2$ ,
\[ \Rightarrow {\text{ 45}} - 21 + 1\]
\[ \Rightarrow {\text{ 25}}\]
Thus there are $25$ straight lines that can be drawn out of $10$ points of which $7$ are collinear.
Hence, the correct option is \[\left( 4 \right){\text{ 25}}\] .

Note: Remember that no line can be drawn through the three non-collinear points. Through one line we can draw infinite lines. And through two points we can draw only one line. Also note that the combination formula is used to find the number of ways of selecting items from a collection as the order of selection does not matter.