nuthan Posted May 28, 2005 Share Posted May 28, 2005 A 4-digits number is consists the numbers selected from 0 to 9 (the numbers should can be used repeatedly, not sure). The number is greater than 1000; the hundreds digit is 0 or 1 and when is 0, the tens' digit cannot be 0. How many such numbers are possible? 9*2*9*10 Quote Link to comment Share on other sites More sharing options...
Jai Gurudev Posted November 15, 2005 Share Posted November 15, 2005 IMO, For the thousands number the choice set is {1-9} = 9choices For Hundreds digit:{0-1} = 2 choices For Tenths digit :{1-9 or 0 -9} based on Hundreds Digit = 9 or 10 choices For Units digit:{0-9} = 10 choices. Here We can deduce Case 1 and Case 2 and add them. Case 1: When Hundreds Digit = 0. So we have, 9*1*9*10 Case 2: When Hundreds Digit = 1 So we have, 9*1*10*10 Adding both we get , 9*1*9*10 + 9*1*10*10 = 9*10(9+10)=90*19 = 1710 Quote Link to comment Share on other sites More sharing options...
CTG1983 Posted November 15, 2005 Share Posted November 15, 2005 I also get 1710. Nuthan, Where were u in last few months? Quote Link to comment Share on other sites More sharing options...
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