Question

Six points in a plane be joined in all possible ways by indefinite straight lines, and if no two of them are coincident or parallel, and no three pass through the same point . (with the expectation of the original 6) . The number of distinct points of intersection is equal to
A.\[105\]
B.\[45\]
C.\[51\]
D.None of these

Answer 1

Hint: In the given question it has asked about the number of possible ways , therefore we have to use permutation and combination . Now , since repetition is allowed , we use combinations which include all possible ways .

Complete step-by-step answer:
As there are \[6\] points and each line segment has \[2\] points , therefore number of lines from \[6\] points , can be calculate using formula \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] ,
where \[{\text{n = number of points}}\] and \[{\text{r = number of points on line segment}}\],
Number of lines from \[6\] points will be \[ = {}^6{C_2}\]
\[ = \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}}\]
On solving we get
\[ = \dfrac{{6!}}{{2! \times 4!}}\]
\[ = \dfrac{{6 \times 5 \times 4!}}{{2! \times 4!}}\]
On further solving we get
\[ = \dfrac{{6 \times 5}}{2}\]
\[ = 15\]
Points of intersection obtained from these lines \[ = {}^{15}{C_2}\]
\[ = \dfrac{{15!}}{{2!\left( {15 - 2} \right)!}}\]
\[ = \dfrac{{15!}}{{2! \times 13!}}\]
On solving we get
\[ = \dfrac{{15 \times 14 \times 13!}}{{2! \times 13!}}\]
\[ = \dfrac{{15 \times 14}}{2}\]
On further solving we get
\[ = 105\]
Now we find the number of times the original \[6\] points come in the intersection . Consider one say \[{A_1}\] joining \[{A_1}\] to remaining \[5\] points we get \[5\] lines and any two lines from these \[5\] lines gives \[{A_1}\] as point of intersection . Similarly , this is the case with the other five points .
Number of times the original \[6\] points comes in the intersection \[ = {}^5{C_2}\]
\[ = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}\]
\[ = \dfrac{{5!}}{{2! \times 3!}}\]
On solving we get
\[ = \dfrac{{5 \times 4 \times 3!}}{{2! \times 3!}}\]
\[ = \dfrac{{5 \times 4}}{2}\]
On solving further we get ,
\[ = 10\]
This for one original points , now for \[6\] points will be \[ = 10 \times 6\]
\[ = 60\]
Also , the \[6\] original points will intersect at least one time , therefore we should also count that .
Now total number of required distinct points of intersection will be \[ = 105 - 60 + 6\]
\[ = 51\]
Therefore the option (C) is the correct answer .
So, the correct answer is “Option C”.

Note: If there are two jobs such that one of them can be completed in \[n\] ways , and when it has been completed in any one of these\[n\] ways , second job can be completed in \[r\] , then the two jobs succession can be completed in \[n \times r\]ways .