how many 4 digit odd no's can be formed such dat every 3 is followed by a 6???

• Case I - 3,6 in first two places, third place has 9 possibilities and last 4 - therefore 1*1*9*4
  
  Case II - 3,6 in 2nd and third place , first digit can be 0-9 excluding 3 and 0, therefore 8*1*1*4
  
  In other cases, odd numbers without digit 3 - 8*9*9*4.
  
  Total = 1*1*9*4 + 8*1*1*4 + 8*9*9*4 = 2660
• 11 years agoHelpfull: Yes(8) No(9)
• if 3,6 are in first two positions then the third can be filled in 10 ways n last digit can be filled in 5 ways(1,3,5,7,9) in order to form a odd no.therefre 1*1*10*5=50(if repetition allowed)
  if 3,6 in 3rd positions thrn 1st position cn b filled in 9 ways n last cn be filled in 5 ways.thrfre 9*1*1*5=45
  total =50+45=95
  
• 11 years agoHelpfull: Yes(1) No(1)
• ans is 2696
• 10 years agoHelpfull: Yes(0) No(0)