a person starts writting four digit numbers, how many times did he write 2

• total 4 digit no. = 9*10*10*10 = 9000
  not containing 2 = 8*9*9*9 = 5832
  total 4 digit number contain 2 = 9000 - 5832 = 3168
  So ans:3168
• 7 years agoHelpfull: Yes(23) No(5)
• lets see from 0-9 --> 1
  from 10-19 --> 1
  from 20-29 --> 11
  from 30-39 --> 1
  from 40-49 --> 1
  from 50-59 --> 1
  from 60-69 --> 1
  from 70-79 --> 1
  from 80-89 --> 1
  from 90-99 --> 1
  
  which makes 20 from 0 to 99
  
  now please observer that similar pattern follows for
  000 - 099 -->20
  100 - 199 --> 20
  300 - 399 --> 20
  400 - 499 --> 20
  500 - 599 --> 20
  600 - 699 --> 20
  700 - 799 --> 20
  800 - 899 --> 20
  900 - 999 --> 20
  
  however for 200 to 299 we have that extra 2 as first digit for all these 100 numbers, so
  200 - 299 --> 120
  
  so
  000 - 999 -->300
  
  so for
  1000 - 1999 -->300
  3000 - 3999 -->300
  4000 - 4999 -->300
  5000 - 5999 -->300
  6000 - 6999 -->300
  7000 - 7999 -->300
  8000 - 8999 -->300
  9000 - 9999 -->300
  
  again for
  2000 - 2999 -->300 + 1000 as we have that extra 2 in the first digit for all these numbers, hence
  
  2000 - 2999 -->1300
  
  So total number of times digit 2 is used 1000 - 9999 is 3700
  
  Join The Disc
• 7 years agoHelpfull: Yes(16) No(0)
• Let the four digit number be abcd.
  
  Case 1: 2 appears only once in the 4-digit number.
  No. of ways of having a number of the form -
  2bcd = 9*9*9 = 729
  a2cd = 8*9*9 = 648
  ab2d = 8*9*9 = 648
  abc2 = 8*9*9 = 648
  
  Total no. of ways = 2673
  Total no. of times 2 appears = 2673
  
  Case 2: 2 appears only twice in the 4-digit number.
  No. of ways of having a number of the form -
  22cd = 9*9 = 81
  2b2d = 9*9 = 81
  2bc2 = 9*9 = 81
  a2c2 = 8*9 = 72
  ab22 = 8*9 = 72
  a22d = 8*9 = 72
  
  Total no. of ways = 459
  Total no. of times 2 appears = 459*2 = 918
  
  Case 3: 2 appears only thrice in the 4-digit number.
  No. of ways of having a number of the form -
  222d = 9
  22c2 = 9
  2b22 = 9
  a222 = 8
  
  Total no. of ways = 35
  Total no. of times 2 appears = 35*3 = 105
  
  Case 4: 2 appears four times in the 4-digit number.
  2222 i.e. only once
  Total no. of times 2 appears = 4
  
  So, total no. of times 2 appears = 2673+918+105+4 = 3700.
• 7 years agoHelpfull: Yes(9) No(0)
• Ans:4000
  X X X X
  for thousand's position no of 2's 1×1000=1000
  for hundredth position no of 2's
  10×100=1000
  for tenth's position no of 2's
  10×10×10=1000
  for one's position no of 2's
  1000
  
  So ans:4000
• 7 years agoHelpfull: Yes(3) No(5)
• use this formula :n*10^(n-1)
  where n is equal to no of digits u want..
• 7 years agoHelpfull: Yes(3) No(0)
• 25 common sence
  
  
• 7 years agoHelpfull: Yes(2) No(7)
• When 2 is at unit place=9*10*10*1
  When 2 is at tenth place=9*10*1*10
  When 2 is at hundred place=9*1*10*10
  When 2 is at thousand place = 1*10*10*10
  So Total no. = 900 + 900 + 900 + 1000 = 3700
• 5 years agoHelpfull: Yes(2) No(0)
• 0 as one girl is already there
• 7 years agoHelpfull: Yes(1) No(1)
• 1 digit---- 1 time
  10---------- (1*10)+10=20
  100---------(20*10)+100=300
  1000--------(300*10)+1000=4000
• 7 years agoHelpfull: Yes(0) No(4)
• total no. ways a 4 digit no can be formed=9*10*10*10
  total no ways in which a 4 digit no can be formed without using 2=8*9*9*9
  total no ways in which a 4 digit no can be formed using 2=(9*10*10*10)-(8*9*9*9)=3168
• 7 years agoHelpfull: Yes(0) No(0)
• i don't know
• 5 years agoHelpfull: Yes(0) No(0)
• When 2 is at unit place=9*10*10*1
  When 2 is at tenth place=9*10*1*10
  When 2 is at hundred place=9*1*10*10
  When 2 is at thousand place = 1*10*10*10
  So Total no. = 900 + 900 + 900 + 1000 = 3700
• 3 years agoHelpfull: Yes(0) No(0)