Home
Class 14
MATHS
There are n points in a plane out of the...

There are n points in a plane out of these points no three are in the same straight line except p points which are collinear. Let "k" be the number of straight lines and "m" be the number of triangles .Then find m−k ?(Assume n=7p=5)

A

`""^(m+n)C_2`

B

`""^(n-m)C_2`

C

`""^(n)C_2-"^(p)C_ 2+1`

D

none of (A) ,(B ) ,(C)

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the difference between the number of triangles (m) and the number of straight lines (k) formed by n points in a plane, given that no three points are collinear except for p points which are collinear. Given: - \( n = 7 \) - \( p = 5 \) ### Step 1: Calculate the number of straight lines (k) The total number of straight lines formed by n points can be calculated using the combination formula for choosing 2 points from n points, which is given by: \[ k = \binom{n}{2} - \binom{p}{2} + 1 \] Where: - \( \binom{n}{2} \) is the number of lines formed by any two points from n points. - \( \binom{p}{2} \) accounts for the lines formed by the p collinear points, which only count as one line. Substituting the values: \[ k = \binom{7}{2} - \binom{5}{2} + 1 \] Calculating \( \binom{7}{2} \) and \( \binom{5}{2} \): \[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] \[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \] Now substituting these values back into the equation for k: \[ k = 21 - 10 + 1 = 12 \] ### Step 2: Calculate the number of triangles (m) The number of triangles formed by choosing any three points from n points is given by: \[ m = \binom{n}{3} - \binom{p}{3} \] Where: - \( \binom{n}{3} \) is the number of triangles formed by any three points from n points. - \( \binom{p}{3} \) accounts for the triangles formed by the p collinear points, which do not form any triangle. Substituting the values: \[ m = \binom{7}{3} - \binom{5}{3} \] Calculating \( \binom{7}{3} \) and \( \binom{5}{3} \): \[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] \[ \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10 \] Now substituting these values back into the equation for m: \[ m = 35 - 10 = 25 \] ### Step 3: Calculate m - k Now we can find \( m - k \): \[ m - k = 25 - 12 = 13 \] ### Final Answer Thus, the value of \( m - k \) is: \[ \boxed{13} \]
Promotional Banner

Topper's Solved these Questions

  • PERMUTATIONS & COMBINATIONS

    ARIHANT SSC|Exercise INTRODUCTORY EXERCISE -(19.6 )|6 Videos
  • PERMUTATIONS & COMBINATIONS

    ARIHANT SSC|Exercise INTRODUCTORY EXERCISE -(19.7 )|12 Videos
  • PERMUTATIONS & COMBINATIONS

    ARIHANT SSC|Exercise INTRODUCTORY EXERCISE -(19.4 )|11 Videos
  • PERCENTAGES

    ARIHANT SSC|Exercise Final round|50 Videos
  • PERMUTATIONS AND COMBINATIONS

    ARIHANT SSC|Exercise HIGHER SKILL LEVEL QUESTIONS|19 Videos

Similar Questions

Explore conceptually related problems

There are 10 points in a plane out of these points no three are in the same straight line except 4 points which are collinear. How many (i) straight lines (ii) trian-gles (iii) quadrilateral, by joining them?

There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. The number of triangles formed by these points, is

There are 10 points in a plane, no three of which are in the same straight line, except 4 points, which are collinear. Find (i) the number of lines obtained from the pairs of these points, (ii) the number of triangles that can be formed with vertices as these points.

Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that that can be formed joining the point.

If three or more points does not lie on the same straight line the points are called -

Out of 18 points in as plane, no three are in the same straight line except five points which re collinear. The num,ber of straight lines formed by joining them is (A) 143 (B) 144 (C) 153 (D) none of these

There are 15 points in a plane, no three of which are in the same straight line with the exception of 6, which are all in the same straight line. Find the number of i. straight lines formed, ii. number of triangles formed by joining these points.

ARIHANT SSC-PERMUTATIONS & COMBINATIONS -INTRODUCTORY EXERCISE -(19.5 )
  1. Find the number of diagonals in a decagon.

    Text Solution

    |

  2. Find the number of diagonals in an n-sided polygon.

    Text Solution

    |

  3. A polygon has 54 diagonals. The number of sides in the polygon is:

    Text Solution

    |

  4. Find the number if triangle that can be formed by joining the...

    Text Solution

    |

  5. Find the number of triangle formed by joining 12 different poin...

    Text Solution

    |

  6. Answer these questions based on the following informations Two paralle...

    Text Solution

    |

  7. There are 3 books of mathematics, 4 of science and 5 of literature. Ho...

    Text Solution

    |

  8. if 20 straight line be drawn in a plane , no two of them bei...

    Text Solution

    |

  9. Find the number of different straight lines obtained by joining n poin...

    Text Solution

    |

  10. There are n points in a plane out of these points no three are in the ...

    Text Solution

    |

  11. There are n points in a plane no three of which are in the same straig...

    Text Solution

    |

  12. If m parallel lines in a plane are intersected by a family of n parall...

    Text Solution

    |

  13. In a plane there are 37 straight lines, of which 13 pass through the p...

    Text Solution

    |

  14. Find the number of different words that can be formed from 15 consonan...

    Text Solution

    |

  15. Find the number of ways of selecting 4 letters from the word EXAMINATI...

    Text Solution

    |

  16. How many words can be formed by using 4 letters at a time out of the l...

    Text Solution

    |

  17. In how many ways can 3 ladies and 3 gentlemen be seated around a ro...

    Text Solution

    |

  18. Eighteen guests have to be seated half on each side of a long table. F...

    Text Solution

    |

  19. There are 4 different letters and 4 addressed envelopes. In how many w...

    Text Solution

    |

  20. There are 5 letters and 5 addressed envelopes.The number of ways in wh...

    Text Solution

    |