Ascent Education Home
• CAT 2016 Classes
• TANCET 2016 Classes @ Chennai
• XAT 2016 Classes @ Chennai
• TANCET Y! Groups
• GRE-GMAT
• US B Schools
• Faculty List
• Question A Day

Permutation and Combination - Quant/Math - CAT 2013

  1. Algebra
  2. Progressions
  3. Averages
  4. Clocks and Calendars
  5. Data Sufficiency
  6. English Grammar
  7. Function
  8. Geometry
  9. Coordinate Geometry
  10. Interest
  11. Mensuration
  12. Mixtures & Alligations
  13. Number System
  14. Percentages
  15. Permutation & Combination
  16. Pipes & Cisterns And Work & Time
  17. Probability
  18. Profit & Loss
  19. Races
  20. Ratio, Proportion
  21. Speed, Time & Distance
  22. Trigonometry
  23. Miscellaneous
  24. General Knowledge
Question 4 the day: May 30, 2003

The question for the day is from the topic of Permutation and Combination.
How many numbers are there between 100 and 1000 such that atleast one of their digits is 6?

(1) 648 (2) 258
(3) 654 (4) 252
Correct Answer - (4)

Solution:


The number of 3 - digit number which have atleast one of their digits 6 can be found by subtracting those 3 - digit numbers which do not have 6 as one of their digits from all the 3 - digit numbers.

We know that the number of 3-digit numbers is 900.

The number of 3 - digit numbers that do not have 6 as one of their digits can be found as follows:

The unit place of the number can be filled in 9 ways (0 to 9, other than 6)

The tenth place of the number can be filled in 9 ways (0 to 9, other than 6)

The hundred's place of the number can be filled in 8 ways (1 to 9, other than 0 and 6)

Therefore, the number of 3 - digit numbers that do not have 6 as one of their digits = 8*9*9 = 648.

Hence, The number of 3 - digit number which have atleast one of their digits 6 = 900 - 648 = 252.




Page top        

 
  © 2002 - 10 ASCENT Education all rights reserved.