Home
Class 14
MATHS
All possible two factors products are fo...

All possible two factors products are formed from the numbers 1, 2, 3, 4, ....., 200. The number of factors out of total obtained which are multiples of 5 is

A

5040

B

7180

C

8150

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the number of two-factor products formed from the numbers 1 to 200 that are multiples of 5, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Range of Numbers**: We are given the numbers from 1 to 200. 2. **Determine the Total Number of Two-Factor Products**: The total number of ways to choose 2 factors from 200 numbers is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \). Here, \( n = 200 \) and \( r = 2 \): \[ C(200, 2) = \frac{200 \times 199}{2} = 19900 \] 3. **Identify Multiples of 5**: We need to find how many of these products are multiples of 5. First, we find how many multiples of 5 are there from 1 to 200. The multiples of 5 in this range are: 5, 10, 15, ..., 200. This forms an arithmetic progression (AP) where: - First term \( a = 5 \) - Common difference \( d = 5 \) - Last term \( l = 200 \) To find the number of terms \( n \) in this AP, we use the formula for the last term: \[ l = a + (n-1)d \implies 200 = 5 + (n-1) \cdot 5 \] Solving this gives: \[ 200 - 5 = (n-1) \cdot 5 \implies 195 = (n-1) \cdot 5 \implies n-1 = 39 \implies n = 40 \] Thus, there are 40 multiples of 5 between 1 and 200. 4. **Determine Non-Multiples of 5**: The total numbers from 1 to 200 is 200. The non-multiples of 5 are: \[ 200 - 40 = 160 \] 5. **Calculate the Number of Two-Factor Products that are Not Multiples of 5**: Now we find the number of two-factor products that are not multiples of 5. This is given by: \[ C(160, 2) = \frac{160 \times 159}{2} = 12720 \] 6. **Calculate the Number of Two-Factor Products that are Multiples of 5**: To find the number of products that are multiples of 5, we subtract the number of products that are not multiples of 5 from the total number of products: \[ \text{Multiples of 5} = C(200, 2) - C(160, 2) = 19900 - 12720 = 7180 \] ### Final Answer: The number of two-factor products that are multiples of 5 is **7180**. ---
Promotional Banner

Topper's Solved these Questions

  • PERMUTATIONS AND COMBINATIONS

    DISHA PUBLICATION|Exercise PRACTICE EXERCISES ( STANDARD LEVEL)|82 Videos
  • PERMUTATIONS AND COMBINATIONS

    DISHA PUBLICATION|Exercise PRACTICE EXERCISES ( EXPERT LEVEL )|48 Videos
  • PERMUTATIONS AND COMBINATIONS

    DISHA PUBLICATION|Exercise TEST YOURSELF|15 Videos
  • PERCENTAGES

    DISHA PUBLICATION|Exercise PRACTICE EXERCISE (TEST YOURSELF)|15 Videos
  • PROBABILITY

    DISHA PUBLICATION|Exercise TEST YOURSELF|15 Videos

Similar Questions

Explore conceptually related problems

All possible two-factor products are formed from the numbers 1, 2,…..,100. The numbers of factors out of the total obtained which are multiple of 3, is

The sum of all natural numbers between 100 and 200, which are multiples of 3 is :

The sum of all possible products taken two at a time out of the numbers +-1,+-2,+-3,+-4 is

Find the total number of factors of 2160 (excluding 1)

The total number of prime factors which are contained in (30)^6 is

All possible numbers are formed using the digits 1, 1, 2, 2, 2 ,2, 3, 4, 4 taker all a time. The number of such numbers in which the odd dighits occupy even places is

If W=2^(4).3^(2).5 then number of factors of W^(2) which are less than W but are not factors of W is:

DISHA PUBLICATION-PERMUTATIONS AND COMBINATIONS-PRACTICE EXERCISES ( FOUNDATION LEVEL)
  1. There are three rooms in a hotel: one single, one double and one for f...

    Text Solution

    |

  2. The digits, from 0 to 9 are written on 10 slips of paper (one digit on...

    Text Solution

    |

  3. The number of ways in which 7 different books can be given to 5 studen...

    Text Solution

    |

  4. In how many ways can 13 different alphabets (a, b, c, ... m) be arrang...

    Text Solution

    |

  5. Number of ways in which the letters of word GARDEN can be arranged wit...

    Text Solution

    |

  6. In how many ways can a mixed double game can be arranged from amongst ...

    Text Solution

    |

  7. In how many ways can 21 identical white balls and 19 identical black b...

    Text Solution

    |

  8. If 5 parallel straight lines are intersected by 4 parallel straight, t...

    Text Solution

    |

  9. The number of ways in which a couple can sit around a table with 6 gue...

    Text Solution

    |

  10. How many different words beginning with O and ending with E can be for...

    Text Solution

    |

  11. How many 6 digit number can be formed from the digits 1, 2, 3, 4, 5, 6...

    Text Solution

    |

  12. There are 5 candidates in an election and 3 of them are to be elected....

    Text Solution

    |

  13. If ""^(2n+1)P(n-1) : ""^(2n-1)P(n) =3 :5 the possible value of n will ...

    Text Solution

    |

  14. All possible two factors products are formed from the numbers 1, 2, 3,...

    Text Solution

    |

  15. A set of 15 different words are given. In how many ways is it possible...

    Text Solution

    |

  16. In an examination, there are 3 multi-choice questions and each questio...

    Text Solution

    |

  17. Seven nouns, five verbs, and two adjectives are written on a blackboar...

    Text Solution

    |

  18. In how many ways can the eight directors, the vice- chairman and the c...

    Text Solution

    |

  19. How many 4 digit numbers divisible by 5 can be formed with the digits ...

    Text Solution

    |

  20. The number of circles that can be drawn out of 10 points of which 7 ar...

    Text Solution

    |