There are $15$ distinct points on a plain sheet of paper. If $4$ of these points are collinear, find the maximum number of triangles that can be drawn using these points.
451
This problem involves finding the maximum number of triangles that can be formed from a set of points, with a specific condition about some points being collinear. We need to select 3 points out of the given points to form a triangle, but we must exclude combinations where the 3 chosen points lie on the same straight line.
Points and Triangle Formation
A triangle is formed by choosing any 3 points that are not on the same straight line (non-collinear). We are given a total of $15$ distinct points on a plain paper.
If all $15$ points were such that no three points were collinear, the total number of triangles would simply be the number of ways to choose 3 points from 15, which is calculated using combinations.
Calculating Total Combinations
The total number of ways to choose 3 points from $15$ distinct points is given by the combination formula: $ \binom{n}{k} = \frac{n!}{k!(n-k)!} $ Here, $n = 15$ (total points) and $k = 3$ (points needed for a triangle).
So, the total number of combinations is:
$ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} $Calculating this value:
$ \binom{15}{3} = 5 \times 7 \times 13 = 455 $This represents the total number of ways to select 3 points from the 15 available points, assuming no three are collinear.
Handling Collinear Points
The problem states that $4$ of these $15$ points are collinear. This means these $4$ points lie on a single straight line.
Any combination of 3 points chosen from these $4$ collinear points will lie on that line and therefore cannot form a triangle. We need to subtract these invalid combinations from our total calculated combinations.
The number of ways to choose 3 points from the $4$ collinear points is:
$ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4 $So, there are $4$ combinations of 3 points that do not form triangles because they are chosen from the set of 4 collinear points.
Maximum Number of Triangles
To find the maximum number of triangles that can be drawn, we subtract the number of combinations that do not form triangles (due to collinearity) from the total number of combinations of 3 points.
Maximum number of triangles = (Total combinations of 3 points) - (Combinations of 3 points from the collinear set)
Maximum number of triangles = $ \binom{15}{3} - \binom{4}{3} $
Maximum number of triangles = $ 455 - 4 $
Maximum number of triangles = $ 451 $
Therefore, the maximum number of triangles that can be drawn using these points is 451.
Important Questions from Permutation and Combination
If Quantity A is the number of ways to assign a number from 1 to 5 without repetition to each of four people, and Quantity B is the number of ways to assign a number from 1 to 5 without repetition to each of 5 people, then which of the following statements is correct with respect to Quantities A and B?
Which of the following muscles regulates the exit of food from the stomach into the small intestine?
There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
In how many ways can we put four different letters into four different envelopes so that atleast three letters go into the wrong envelopes?
