From digit 0-9, how many no. can be made such that two adjacent digits are same. like-119,911.....

• 112,113-119=9
  220,221,223-229=9
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  990-998=9 so total 9*9=81(consecutive digits at fast)
  122,133,144,155,166,177,188,199=8
  211,233,244,255,266,277,288,299=8
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  911,922,933,944,955,966,977,988=8 so total 8*9=72(consecutive digits at last)
  Now 111,222,333,444,555,666,777,888,999 so total=9(all digits same)
  Again 100,200,300,400,500,600,700,800,900 so total=9(consecutive two zeroes at last)
  
  So al total two consecutive digits same= 81+72+9+9=171(Ans)
• 10 years agoHelpfull: Yes(36) No(2)
• 11X & X11 = 10 + 9 = 19
  similarly 22X & X22 .. 33X & X33 upto 99X & X99
  So,(10+9)*9=171
• 10 years agoHelpfull: Yes(14) No(1)
• i think 171 is the ans
  
• 10 years agoHelpfull: Yes(8) No(3)
• let us consider three digit no --- say first two filled with 11 and third can be filled in 10 ways so ... similarly with starting 22 can be filled in 10 ways..
  it gives 9*10=90 ways
  
  then other possible is --- filling the 2nd and 3rd place with same no this can be done in 10 ways and first place with 9 ways because 0 we can place at hundred place.. it gives 9*10=90ways
  so total =90+90=180ways but we want to subtract 9 ways because in 1st and 2nd possible we counted the numbers 111 222 333 444 555 666 777 888 999 two times so we are going to subtract 9 ways ....
  final total no of ways are 180-9=171 ways
  
  
• 10 years agoHelpfull: Yes(6) No(1)
• if only two adjacent digits need to be same then.. ans is 162.
  if all three digits can be same.. then ans is 171
  
• 10 years agoHelpfull: Yes(4) No(0)
• @ANKITBALYAN ans ki maa ki aakh , Solution nahi aata kya.
• 10 years agoHelpfull: Yes(4) No(0)
• no. containing 2 1's
  11(2-9)=16 [two 1's is fixed & 3rd digit may be from 2-9]
  similarly for 22,33,..., 99 for each 16 no. will be formed
  
  no. containing 00 => 9(100,200,...,900)
  
  total=9*16+9=153
  
• 10 years agoHelpfull: Yes(3) No(12)
• @ rakesh u have taken 11(2-9)...and what about 110,220,330.....
  
• 10 years agoHelpfull: Yes(1) No(0)
• let the no. is abc
  1st find unfavourable no.
  a=1,2,3.......,9=9 possibilities(except 0)
  b=9 possibilities(except a)
  c=9 possibilities(except b)
  Thus, possible unfavourable no.= 9*9*9=729
  Total no=900
  Thus, Favourable no=900-729=171
• 9 years agoHelpfull: Yes(1) No(0)
• qustn should have been given clearly like 2 digit number or 3 digit number or 4 digit number...
• 9 years agoHelpfull: Yes(1) No(0)
• then what about 111,222,333,444....999?? these nos are also there.
• 10 years agoHelpfull: Yes(0) No(0)
• Can someone pls explain how its 171..
• 10 years agoHelpfull: Yes(0) No(2)
• 110-119=>10,211-911=>9 So for each digit there will be 19 possibilities
  
  so 19*9=161;
  
  & for digit 0=> 0 cant be in 100s position,so only 10 combinations will be possible. totally 161+10:171 combinations are possible
• 10 years agoHelpfull: Yes(0) No(1)