howmany 3 digit numbers are there tens digit place is more than hundreds digit place and units place less than hundreds digit place?

• three digit numbers between that can be made are between 100 to 999
  1. consider no 1_0 where hundreth place is greater than units place but for tens digit to be greater possible
  combinations are tens digit can be 2,3,4,5,6,7,8,9 hence 8 combinations.
  2.Now consider no 2_0,2_1 where hundreth place is greater than units place but for tens digit to be greater possible
  combinations are tens digit can be 3,4,5,6,7,8,9 hence 2*7=14 combinations.
  3.Similarly for no 3_0,3_1,3_2 there will be 18,for no 4_0,4_1,4_2,4_3 there will be 20, for no 5_0 to 5_4 there will be again 20 combinations and so on...
  4. Thus order followed will be
  8
  14
  18
  20
  20
  18
  14
  8
  thus in all 120 combinations. Hence 120 is the answer.
  
  
• 8 years agoHelpfull: Yes(23) No(0)
• 1+3+6+10+15+21+28+36=120
  By not using Permutation and Combination.
  120
  231 230 130
  342 341 340 241 240 140
  453 452 451 450 352 351 350 251 250 150
  ..................................................................................................and so on
  I didn't think i can apply P&C so i interpret the algorithm.
• 8 years agoHelpfull: Yes(7) No(1)
• 9C3+9C2=120
• 8 years agoHelpfull: Yes(4) No(5)
• keep the tens place fix.
  start from 9. _9_ = possible options 1-8 on hundred place and 0-7 on unit place. so 8*1*8
  now _8_= Same 1-7for hundred and 0-6 for unit . so 7*1*7
  so on _7_ = 6*1*6
  _6_ = 5*1*5
  _5_= 4*1*4
  _4_= 3*1*3
  _3_ = 2*1*2
  _2_ = 1*1*1
  Finally we get 8*8+ 7*7+ 6*6 + 5*5+ 4*4+ 3*3+ 2*2+ 1*1 = 204
• 8 years agoHelpfull: Yes(4) No(4)
• If unit place is 0 and by putting 9,8,7,6,5,4,3,2 at tens place we will get
  8 comb, 7 comb, 6 comb, 5 comb, 4 comb, 3 comb, 2 comb, 1 combination respectively
  = 8+7+6+5+4+3+2+1 = 36
  Similarly by putting 1 at unit place and putting 9,8,7,6,5,4,3 at tens place we will get
  7+6+5+4+3+2+1 = 28 combinations
  Similarly by putting 2 at unit place and putting 9,8,7,6,5,4 at tens place we will get
  6+5+4+3+2+1 = 21
  Similary we will get
  5+4+3+2+1=15
  4+3+2+1=10
  3+2+1=6
  2+1=3
  1=1
  Total = 36+28+21+15+10+6+3+1 = 120
• 7 years agoHelpfull: Yes(3) No(0)
• 8*1+7*2+6*3+5*4+4*5+3*6+2*7+1*8=120
• 8 years agoHelpfull: Yes(2) No(2)
• ans is 204
  
  For solving this first fix the middle number and then check for the option left for ones and hundred digit.
  Also we cannot have 1 & 0 in tens place. so start fixing the number from 2 onwards.
  Then we will get the series as :
  
  1*1+2*2+3*3+4*4+5*5+6*6+7*7+8*8 = 204
  
• 8 years agoHelpfull: Yes(2) No(8)
• Read the question carefully xyz is a 3 digit number then xz satisfies these digits
  890-897=8
  780-796=17
  670-695=26
  560-594=35
  450-493=44
  340-392=53
  230-291=62
  120 then total 246 numbers
• 7 years agoHelpfull: Yes(1) No(1)
• 120 is correct answer no dout..
• 7 years agoHelpfull: Yes(1) No(1)
• how to prepare for elitmus ??
• 8 years agoHelpfull: Yes(0) No(1)
• answer is
  120
• 8 years agoHelpfull: Yes(0) No(0)
• lets take a 3 digit number is abc where a>b>c ,
  if i take ab as single entity of pairs likes (1,9) than c can only be 0 that is one possible
  similarly (2,9)={0,1}.............(8,9)={0,1,2,3,4,5,6,7,8}=8 possiblities
  and the sum of
• 8 years agoHelpfull: Yes(0) No(1)
• We need to form 3 digit number,
  _ _ _ the rule says that tens should be greater than hundredth and it should be greater than unit digit.
  So overall Tenth(T)>Hundredth(H)>Unit(U).
  if we will can begin with middle one as it will be easiest.
  Note** We cant take H digit to be 0, it wont be 3 digit number.
  
  Min value H can take is 2, so T should be less than 2 i.e. 1 or 0(we can take 0 at T), but since T>U, T=1 and finally U=0. Number is 120.
  
  Similarly, for H=3 , T= 2 or ! and U= 1 or 0 resp. So, numbers would be 231,230,130. So on,
  it is a cumulative addition,
  
  Addition= previous_addition + n-1
  Ex. for n(i.e. H) = 3 ,
  Addition = 1+ (3-1)
  = 1+ 2 = 3 ways.
  For, H=4 , Addition = 3 + 3 = 6 and so on.
  
  so, total = 1 + 3 + 6 +10 +15 +21 +28+ 36 = 120 ways.
• 7 years agoHelpfull: Yes(0) No(0)