Question

Find the number of straight lines obtained by joining 10 points on a plane if.
A. No three of which are collinear.
B. Four points are collinear.

Answer 1

Hint: According to the question given in the question new have to determine the No three of which are collinear and if four points are collinear when the number of straight lines obtained by joining 10 points on a plane. So, first of all we have to determine the number of straight lines obtained by joining 10 points on a plane if no three points are collinear which can be obtained as mentioned in the question that the straight line is formed by joining the 10 points in which we have to take no three which are collinear.
Now, we have to determine the events for the case and then we have to use the permutation and combination to choose the favourable points out of the total points which can be said all the out of the total number of outcomes with the help of the formula as mentioned below:

Formula used:
$ \Rightarrow c_r^n = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}...............(A)$
So with the help of the formula above we can determine the required number of ways.
Now, we have to determine when Four points are collinear, so that there will be 4 favourable events because a line is formed by two points. Then we have to use the formula (A) as mentioned above to determine the required number of ways.

Complete step by step answer:
Step 1: First of all we have to determine the number of straight lines obtained by joining 10 points on a plane if no three points are collinear which can be obtained as mentioned in the question that the straight line is formed by joining the 10 points in which we have to take no three which are collinear as mentioned in the solution hint.
We have to choose the 2 points and as we know that total number of ways or sample space is 10.
Step 2: Now, we have to use the formula (A) as mentioned in the solution hint to determine the required number of ways hence, on substituting all the values in the formula (A),
$
   \Rightarrow c_2^{10} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} \\
   \Rightarrow c_2^{10} = \dfrac{{10!}}{{2!8!}} \\
   \Rightarrow c_2^{10} = \dfrac{{10 \times 9 \times 8!}}{{2! \times 8!}} \\
   \Rightarrow c_2^{10} = 45 \\
 $
Step 3: Now, we have to determine for when Four points are collinear so, for that there will be 4 favourable events because a line is formed by two points.
Step 4: Now, we have to use the formula (A) as mentioned above to determine the required number of ways as mentioned in the solution hint. Hence,
$
   \Rightarrow c_2^4 = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \\
   \Rightarrow c_2^4 = \dfrac{{4!}}{{2!2!}} \\
   \Rightarrow c_2^4 = \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}} \\
   \Rightarrow c_2^4 = 6 \\
 $
Step 5: Now, with the help of the solution step 4 we can determine the number of ways is four points are collinear as,
$
   = c_2^{10} - c_2^4 + 1 \\
   = 45 - 6 + 1 \\
   = 46 - 6 \\
   = 40 \\
 $

Hence, with the help of the formula (A) as mentioned in the solution hint we have determined the required number of ways when No three of which are collinear is 45 and when Four points are collinear is 40.

Note: To determine the number of ways for a required event first of all we have to choose the required event from the given event then we have to choose this event from all possible outcomes.
To determine the probability we have to divide the favourable event by the sample space or total number of outcomes.