How many ways the three digits number (XYZ) can be formed in the form XZ?

• Submitted by nishant:- take value of X Y Z in 0 -9
  here value of x cant be 0 as its in cent place.
  
  take value of x - 1 to 9
  y -- 0 -9
  z -- 0 - 9
  
  now put x =1 y = 0 then z = 0 to 9 here we cant use 1 for y and 1 and 0 for z.
  1 0 2, 10 3, 10 4, 105, 106, 107, 108, 109
  so for x = 9 value possible and for z = 8 ---so total value possible are 72 and in y we can put different 8 values so answer is 9 * 8 * 8 == 64*9= 576
  
• 8 years agoHelpfull: Yes(11) No(6)
• Total number of 3 digit numbers = 9×10×10 = 900
  Total number of numbers in which no digit repeats = 9×9×8 = 648
  So the total number of numbers in which at least one digit repeats = 900 - 648 = 252
• 7 years agoHelpfull: Yes(3) No(0)
• Above question is wrong right question is
  
  A three digit number of the form ‘xyz’ is to be formed such that x > y, x > z, and y < z. How many such numbers are possible if x > 0?
• 7 years agoHelpfull: Yes(2) No(0)
• i think question is wrong
  if x
• 7 years agoHelpfull: Yes(1) No(1)
• data insufficient
• 7 years agoHelpfull: Yes(1) No(0)
• x=1
  y=2 z= 3...9 (7 ways)
  y=3 z= 4...9 (6 ways)
  ...
  y=8 z= 9 (1 way)
  
  Total : 7+6+...+1
  
  x=2
  y=3 z= 4...9 (6 ways)
  y=4 z= 5...9 (5 ways)
  ...
  y=8 z= 9 (1 way)
  
  Total : 6+5+...+1
  
  ...
  
  x=6
  y=7 z= 8,9 (2 ways)
  y=8, z=1 (1 way)
  Total : 2+1
  
  x=7
  y=8 z= 9(1 way)
  
  Total : 1
  
  
  Required number of ways = ( 7+6+...+1) + (6+5+...+1) + (5+4+3+2+1) + (4+3+2+1) + (3+2+1) + (2+1) + 1
  =28+21+15+10+6+3+1
  = 84
  Qn is x
• 7 years agoHelpfull: Yes(1) No(0)
• I think difference shud be 11 only
• 8 years agoHelpfull: Yes(0) No(0)
• 3p2
  i.e
  only 6 i think
• 7 years agoHelpfull: Yes(0) No(0)
• assuming x cannot be zero
  xyz
  x,y,z can be 0,1,2,3,4,....9
  1)y=0 not possible
  1)y=1 x=z=0 not possible
  2)y=2 x,z can be filled with{0,1} =1*2= 2case..............{x in 1 way and z in 2{x cannot be zero}}
  3)y=3 x,z can be filled with{0,1,2}=2*3=6 case
  ..
  ..
  9)y=9 x,z can be filled with{0,1,2..8}=8*9=72 cases
  
  Total cases 1*2+2*3+..........n(n+1)=1+n(n+1)(n+2)/3
  here n is 8
  hence total cases=240 ans
  
  if x can be zero
  than no of two digit number with y>z is 9{10,20,30..90}+8{21,31,41..91}+7{32,42....
  
  hence total case=285
• 7 years agoHelpfull: Yes(0) No(0)