i started writing 4 digit number, how many times i had written 2???/

• option (d) this is absolutely correct
  total no. of 4 digit nos are 9000(b/w 1000 and 9999).
  consider the case where the nos contain no 2 in it.
  so units place can be filled in 9 ways(0,1,3,4,5,6,7,8,9)
  similarly,tenth place in 9 ways
  hundredth place in 9 ways
  thousands place can be filled in 8 ways only(because 0 cannot come)
  so total nos are (9000-(8*9*9*9))=3168
• 10 years agoHelpfull: Yes(30) No(4)
• if we take 2 at unit place then we can arrange the digits at thousand place with 9 ways(1,2,3,4,5,6,7,8,9)
  lly, at hundred place we can arrange the digits in 10 ways(0,1,2,3,4,5,6,7,8,9)
  lly,at tenth place we can arrange the digits in 10 ways(0,1,2,3,4,5,6,7,8,9).
  so we have (9*10*10)ways= 900ways
  
  in second case taking 2 at tenth place we have
  Number of 2‘s at tenths place(1000 to 9999)=900
  
  in third case taking 2 at hundredth place we have
  Number of 2‘s at hundreds place(1000 to 9999)=900
  
  in fourth case taking 2 at thousands place we have
  Number of 2‘s at thousands place(1000 to 9999)=1000
  
  therefore total number of 2‘s =(900+900+900+1000)=3700ans
• 10 years agoHelpfull: Yes(18) No(8)
• If Question is How many times we will write 2 then answer is 3700
  And, If Question is how many number have the number 2 in it then answer is 3168
• 10 years agoHelpfull: Yes(10) No(1)
• as a four digit number the possibility of 2 at unit digit becomes 10*10*9 , because we can arrange the numbers as as from 0 to 9.
  similarly as from the tenth digit the possibility of getting 2 becomes 10*10*9 which becomes again 900,
  as same for the hundreth place the number becomes 900,
  now coming for the thousands place the number becomes 10*10*10,
  the total becomes 900+900+900+1000=3700
• 10 years agoHelpfull: Yes(5) No(7)
• 3168
  9000-5832
• 10 years agoHelpfull: Yes(4) No(6)
• 2 at unit place from (1000-9999)=900
  2 at tens place from (1000-9999)=900
  2 " hundredth place from (1000-9999)=900
  " " thousand place (1000-9999)=1000
  
  total=900+900+900+1000=3700
• 10 years agoHelpfull: Yes(2) No(1)
• 2 occurs 3168 tymes in 4 digit numbers
• 10 years agoHelpfull: Yes(0) No(3)
• 1000 times(Ans)
  Explanation-
  Number is of the form (2---)
  1st place only 2 can come
  2nd place 10 digits can come
  3rd place 10 digits can come
  4th place 10 digit can come
  hence 10*10*10=1000
• 10 years agoHelpfull: Yes(0) No(8)
• ans=2080
  no.of 2 in unit place=90
  no.of 2 in tens place=90
  no.of 2 in hundred place=900
  no.of 2 in thousand place=1000
  hence total no.of 2= 90+90+900+1000=2080
• 10 years agoHelpfull: Yes(0) No(0)
• Ans.after 2 we have 3 places, in this 3 places we can put digit from 0 to 9 means total 10 digit.
  10*10*10=1000 using permutation and combination
  alternative way====
  from starting with 2 we have number 2000 to 2999 because given first digit must be 2 so
  according AP first a=2000 last number l=2999 and d=1 then
  l=a+(n-1)d
  2999=2000+(n-1)1
  2999-2000=n-1
  999+1=n
  ans n=1000
• 10 years agoHelpfull: Yes(0) No(0)
• please anyone explain and write solution of this question as soon....
• 10 years agoHelpfull: Yes(0) No(0)
• there is possibility of coming no.2222 in all the four cases then the same no. 2 will be counted 4 times extra..how to egnore it?? @PUJA
• 10 years agoHelpfull: Yes(0) No(0)
• the ans is
  9*10*10*1=900
  9*10*1*10=900
  9*1*10*10=900
  1*10*10*10=1000
  
  3700
  
  
• 9 years agoHelpfull: Yes(0) No(0)