A three digit number of the form ‘xyz’ is to be formed such that x z. How many such numbers are possible ?

• So.....if y=0 then following cases would exist..
  For x=2, 201
  For x=3, 301, 302
  For x=4, 401,402, 403
  For x=5, 501,502,503,504
  For x=6, 601,602,603,604,605
  For x=7, 701,702,703,704,705,706
  For x=8, 801....... 807
  For x=9, 901,902,903 .......908
  
  So total count with y=0 is 1+2+3+4+5+6+7+ 8= 36
  
  And the total count done by akash=84
  
  So the answer is 84+36= 120
• 8 years agoHelpfull: Yes(7) No(0)
• ur question is incomplete...its like,,",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?"
  
  for solution....see 20th question in the following link.
  http://www.campusgate.co.in/2011/11/permutations.html
• 8 years agoHelpfull: Yes(4) No(1)
• 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?
  
  x=0 && x=1 && x=2 is not possible because y and z has to be smaller.so for x=3 total no-=1
  x=4, 412 413,423..... total no=3
  x=5 512,513,514,523,524,534 total no=6
  x=6 612,613,614,615,623,624,625,634,635,645=10
  x=7 712,713,714,715,716,723,724,725,726,734,735,736,745,746,756 total no-15
  x=8 812,813,814,815,816,817,823,824,825,826,827,834,835,836,837,845,846,847,856,857,867 total no=21
  x=9,............ total no 28.
  Add all 1+3+6+10+15+21+28=84
  
• 8 years agoHelpfull: Yes(3) No(0)
• Akash...this solution is infact wrong....coz it doesn't include some cases ...eg. value of y could be zero and hence the possible no. Are 301, 302, ...
  
  
  403,402, 401 etc..since there is no restriction of non zero number
• 8 years agoHelpfull: Yes(3) No(0)
• x=2, 201 1
  x=3,301 302,312 3
  it will form a series i.e
  1+(1+2)+(1+2+3)+............+(1+2+3+4+5+...+8)=167ans
  
• 7 years agoHelpfull: Yes(0) No(0)
• Ans is 120
  include y = 0 also
• 7 years agoHelpfull: Yes(0) No(0)