How many 3 digit numbers can be formed using digits 1,2,3,4 and 5 without repeatation, such that number is divisible by 6.
(a) 4 (b) 6 (c) 8 (d) 10

• 8 is ans
  312,132,432,342,324,234,534,354
  
• 9 years agoHelpfull: Yes(18) No(1)
• 8. Is the correct answer
• 9 years agoHelpfull: Yes(3) No(1)
• if the no is to b divsible by 6 then it should also divisible by 2 and 3
  
  for 2 the rule is last no should b even so here v have 2 nd 4
  so thn no vll lyk this __2 nd __4
  
  and the no should not to b repeated so now v cannt use 2 nd 4 , so now v have 1,3,4,5 for no ending vth 2 nd 1,2,3,5 for no ending vth 4
  for divisble vth 3 the rule is total of that no should b mutiple of 3 for no ending vth 2 means v can have 312,132,432,342
  nd for no ending vth 4 means 324,234,534,354
• 9 years agoHelpfull: Yes(3) No(3)
• as the no have to b divisible by 6 then it should be divisible by 2 and 3 both....that means no. have to be even as well as the sum of the no. should be divisble by 3..
  
  so...we have 2 and 4 for the unit place.....
  
  now for _ _ 2 we have 4 combinations as 2 has been used we cant use this....so possible combo will be 1,3 or3,1or4,3or3,4....i.e.....=4 nos...
  
  now for _ _ 4 we have 2,3 or 3,2 or 3,5 or 5,3...i.e= 4nos....
  
  so total nos will be=8
• 9 years agoHelpfull: Yes(2) No(1)
• ans>8
  312,132,432,342,324,234,534,354
• 9 years agoHelpfull: Yes(1) No(1)
• possible numbers are 132,312,642,462,162,612,432,342, 234,324,624,264,564,654,534,354. BUT 16 is not given in option.
• 9 years agoHelpfull: Yes(1) No(14)
• How 8 is possible
• 9 years agoHelpfull: Yes(0) No(0)
• please explain the answer
• 9 years agoHelpfull: Yes(0) No(1)
• 8 is ans
  312,132,432,342,324,234,534,354
• 9 years agoHelpfull: Yes(0) No(1)
• 5C3 => 5C2 [ ncr = ncn-r]
  so 5*42*1 = 20/2
  i.e 10 ans will be 10
• 9 years agoHelpfull: Yes(0) No(5)
• ans - 10
  324, 534 312 612 342
  234 354 132 162 432
• 9 years agoHelpfull: Yes(0) No(4)