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 September 2nd, 2019, 10:55 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 486 Thanks: 1 How to solve this problem? How many numbers lie between 300 & 500 in which 4 comes only 1 time?? Pls explain this problem step by step. Is this problem belongs to permutation topic??
 September 2nd, 2019, 11:50 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1475 Suppose the 4 is in the first digit. The second digit and third digits can be 0-3,5-9. Thus there are $9^2=81$ numbers in range with the first digit being the only 4. Suppose it is in the 2nd digit. The first digit must be 3. The 3rd digit can be anything but 4. So there are 9 numbers with the only 4 as the second digit. The situation is identical if 4 is the 3rd digit and thus there are 9 numbers with the only 4 being the third digit. Summing these we get 99 different numbers 300-500 with 1 4 in them.
 September 3rd, 2019, 07:49 AM #3 Senior Member   Joined: Aug 2014 From: India Posts: 486 Thanks: 1 How you got 9^2 in second sentence?
September 3rd, 2019, 10:50 AM   #4
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Quote:
 Originally Posted by Ganesh Ujwal How you got 9^2 in second sentence?
2 digits of 9 choices each

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