In how many ways can the digits 2,3,5,7 and 9 be placed to form a three-digit number so that the higher order digit is always greater than the lower order digits?

1: 8
2: 9
3: 10
4: 15

• Ans : 10 ways
  
  Given
  100th digit is always greater than the 10's digit nd unit digit
  
  so possible numbers for 100th position are 9, 7 ,5
  if we select 3, d remaining 2 can occupy 2, 5 ,7, 9 which is greater than the 100th digit...
  
  so if d 100th digit is "9" remaining 2 digits may any 2 among 4 {2 ,3 ,5 ,7} --> 4C2 = 6
  so if d 100th digit is "7" remaining 2 digits may any 2 among 3 {2 ,3 ,5} --> 3C2 = 3
  so if d 100th digit is "5" remaining 2 digits may any 2 among 2 {2 ,3} --> 2C2 = 2
  
  so totally 6 + 3 + 1 = 10ways
  
  Ans : 10
• 10 years agoHelpfull: Yes(139) No(4)
• 3)10
  
  The 3 digit numbers are:
  975 ,973,972,953,952, 932,753,752,732,532
• 10 years agoHelpfull: Yes(23) No(1)
• answer should be 10 if we take 9 in place of 100 then it will b
  4c2=6
  if we take 7 as 100 then it will be 3c2 ie 3
  if we take 5 as 100 then it will 2c2 ie 1
  so answer is 6+3+1=10
• 10 years agoHelpfull: Yes(9) No(1)
• sorry 2C2 = 1
• 10 years agoHelpfull: Yes(5) No(1)
• 9 7 (2 3 5) 3 ways
  9 5 (2 3) 2 ways
  7 5 (2 3) 2 ways
  5 3 2 1 way
  9 3 2 1 way
  7 3 2 1 way 10 is answer
• 10 years agoHelpfull: Yes(3) No(0)
• its solve practically;
  975,973,972,953,952,932;
  753,752,732;
  532;
  so 6+3+1=10
  
• 10 years agoHelpfull: Yes(1) No(0)
• 10 ways :D
• 10 years agoHelpfull: Yes(1) No(0)
• since hu>ten>unit
  5*4*3/3!=10
  _ _ _
• 10 years agoHelpfull: Yes(1) No(0)
• 5P3/3!=10
  here 3! indicates the arranged 3 digits should be in order.
• 8 years agoHelpfull: Yes(1) No(0)
• its very simple 5c3 because we are not permuting the no. . so a particular combination will have only one kind of arrangement .
• 7 years agoHelpfull: Yes(1) No(1)