Find the total no of 3 digits number such that if one of the digit is 3 then it must be followed by 7.

• two forms is possible
  _37 or 37_
  
  _37 => _ can be 1,2,4,5,6,8 9 => 7 no.s
  
  37_ => _ can be 0,1,2,4,5,6,8,9 => 8 no.s
  
  total= 7+8 =15
  
• 10 years agoHelpfull: Yes(40) No(9)
• @rakesh question says "if" what about the other cases where the digit 3 is not present. I think we have to consider them also. because from what i remember 15 was not among the options . and yea repetitions are not allowed which u already assumed
• 10 years agoHelpfull: Yes(10) No(0)
• Two possibilities _37 or 37_
  _37=> 8 ways
  37_=> 9 ways
  
  the numbers that don't have 3 as a digit must also be considered.
  8*9*9 => 648 numbers that don't have 3 as a digit.
  
  So total numbers are 648+8+9= 665.
• 10 years agoHelpfull: Yes(9) No(3)
• i think 19 =>137,237,337,437,537,637,737,837,937 total 9
  and 370,371,372,373,374,375,376,377,378,379 =>10
  10+9=19
  
• 10 years agoHelpfull: Yes(4) No(12)
• if 3 is not present
  then, 8*8*7=448
  and if 3 is present
  then, 6*1*1=6
  or 1*1*8=8
  so total=448+6+8=462
• 10 years agoHelpfull: Yes(2) No(1)
• total no. of 3 digit no.having each digit distinct are
  8*8*7=468
  and now
  3 digit no.'s having one of d digit 3 and followed by 7=15 where each digit is distinct.
  so total no. of such type of no.'s are =468+15=483
• 10 years agoHelpfull: Yes(2) No(0)
• i think the answer is 15 as if first position is 3 then 2nd is 7 and third can be filled in 8 ways excluding 3&7. and in case 2,2nd and 3rd position is 3&7 and first position can be filled in 7 ways excluding 0,3&7. so total ways=8+7=15
• 10 years agoHelpfull: Yes(1) No(0)
• If it is said that Repetition is not allowed then there are two cases 1st) When 3 is at the beginning,2nd)When 3 is at the middle
  For 1st case 3_ _ there are 8 possible ways
  For 2nd case _3_ there are 7 possible ways
  Therefore total there are 15 ways
• 10 years agoHelpfull: Yes(1) No(0)
• i think we should consider the numbers also like 731.. bcaz the question is not bounded to 7 followed by 3
• 10 years agoHelpfull: Yes(1) No(1)
• ans:27
  137,237,337,437,537,637,737,837,937 total 9
  370,371,372,373,374,375,376,377,378,379 =>10
  and we can consider 3_7 also bcoz here not mentioned immideatly followed so
  i think we can consider 307,317,327,337,347,357,367,387,397 total=8
  totaal=27
• 10 years agoHelpfull: Yes(0) No(2)
• 3 must be followed by 7 but 7 can be followed by any no. except 3
  possibility _37 or 37_
  137, 237, 437, 537, 637, 737, 837, 937 = 8
  370, 371, 372, 374, 375, 376, 377, 378, 379 = 9
  total= 8+9 =17
  not possible cases are 037,337, 373
• 10 years agoHelpfull: Yes(0) No(3)
• rakesh wat abt 3rd possibility 73_ it wont be occur r else wat
• 10 years agoHelpfull: Yes(0) No(1)
• 2 forms possible :
  _37 and 37_
  1st case _37, 1,2,4,5,6,7,8,9 can come as 3 is folled by 7 but evry 7 doesnt follow 3 so 7 will b considered.i.e. 8 possibilities
  2nd case also 37_ 0,1,2,4,5,6,7,8,9 9 possibilities
  so..8+9=17
  
• 10 years agoHelpfull: Yes(0) No(0)
• To fill first position of 3 digit number we have eight possibility i.e 8ways
  2nd position fill with 7 ways
  3rd position with 6 ways
  
  Now next condition if 3 occur in number then 7 is also came in that no. So at end we have 7 ways
  To fill last position.
  (8×7×6)+7=343.
• 10 years agoHelpfull: Yes(0) No(0)
• (9p3-8p2)(for the case not having 3)+8( for 37_)+7(for _37)=463
• 10 years agoHelpfull: Yes(0) No(0)
• We need to consider two cases:
  i> when 3 is there in the no., then rules follow
  ii> when 3 is not there, we use simple permutation rules for counting.
  Ans. will ne result of & .
  
  Moreover, question is not clear whether we need to count the number where 3 is at the units place.
• 10 years agoHelpfull: Yes(0) No(1)
• not an elitmus level question...
• 10 years agoHelpfull: Yes(0) No(6)