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1. 
   neela Teacher
   Graduate School

   eNotes Editor

   Three digit natural numbers are 100 to 999 and the number of numbers is 999-100+1= 900.

   Hundred's place could be occupied by any number 1 to 9 . That is 9 ways.

   Having filled up the 1st place , the 2d  place has tobe filled by any of the 9 ways  leaving the digit at 1st place similarly the third place can be filled by  8 digits left out after 1st and second diglaces. So the number of different 3 digit number is 9*9*8=648. Therefore, there are 900-648=252 numbers which are not different digits (inclusive of 2 digits-equal number and  all 3 digit-equal numbers)

   Note that the 252 ways includes the all three digits being equal in 9 ways, 0 not bing possible in the 100s place. like:111,222,333,....999.

   Thererefore, the required probablity = possible number of ways favouring two equaldigits/all possible ways of 3 digit number

   =252/900=0.0.28 , including all three equal digits possiblity.

   The probability of getting only two equal digits numbers = (252-9)/900 =243/900 = 0.27.

    

    

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   Posted by neela on Saturday June 20, 2009 at 9:25 AM
   
   

2. 
   giorgiana1976 Teacher
   Doctorate

   eNotes Editor

   For finding the numbers formed by unlike digits we have to apply the following formula:

   We'll subtract arrangement of 9 digits, taken 2 of them from arrangement of 10 digits, taken 3 of them.

   arrangement of 9 digits, taken 2 of them = 9!/(9-2)!=9!/7!=

   =7!*8*9/7!=72

   arrangement of 10 digits, taken 3 of them=10!/(10-3)!=10!/7!=7!*8*9*10/7!=720

   Arr10,taken 3- Arr9,taken 2= 720-72=648

   Numbers with all digits equal between them are 9.

   Numbers formed by 3 digits and 2 of them equal=

   =900-648-9=243

   The probaility P=243/900=27/100

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   Posted by giorgiana1976 on Saturday June 20, 2009 at 9:26 AM