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Permutation and Combination - Quant/Math - CAT 2009
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 |
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(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.
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