How many six digit numbers can be formed using digit 1 to 6, without repition such that the number is divisible by the digit in it's unit place

a.402
b.528
c.648
d.720

• First we fix 1 at unit place of a six digit number.
  Since any number can be divided by 1 .
  So it does not matter what comes at other five places .
  _ _ _ _ _ 1
  So, first position can be filled in 5 ways as any number from 2 to 6 can be placed in that position . Similarly 2nd position can be filled with 4 ways .
  So , by fundamental principle of counting ,
  We get,
  
  5x4x3x2x1x1=120 or 5! x 1=120
  
  Now for 2 at units place ,
  
  Any even number is divided by 2 ,
  Hence it is again similar to first case,
  5! × 1 =120
  
  Now for 3 at units place ,
  
  For any number to be divisible by 3 , we need to make sure that sum of the digits is divisible by 3.
  Now , 1+ 2 + 3 +4 + 5 +6 = 21
  Since 21 is divisible by 3 , so any number formed by number 1 to 6 is divisible by 3 .
  Hence , 5! × 1 =120.
  
  Similarly for 5 , 5! ×1 = 120
  
  Now for 6 at unit place ,
  We know that any number which is divisible by 2 and 3 is also divisible by 6 .
  
  Hence , 5! ×1 =120
  
  Now for 4 at unit place.
  If there is even number at tens place with 4 at units place , then the number is divided by 4 else no.
  
  So we have only two even number 2 and 6 which we can put at tens place .
  So now first fixing 2 at tens place and 4 at units place .
  _ _ _ _ 2 4
  
  We get by fundamental principle of counting ,
  4 × 3 × 2 × 1 × 1 ×1
  Or
  4! × 1 ×1 = 24
  
  Similarly for 6 at tens place and 4 at units place .
  We get 4! × 1 × 1 = 24
  
  Now let us add all the values we got ,
  
  
  120 +120 +120 + 120 +120 +24 +24 = 648
  
  Hence answer is (c)
• 7 years agoHelpfull: Yes(32) No(5)
• 1 units place divisable by 1 so 5!=120
  2 units place is divisable by 2 so 5!=120
  3 units place is divisable by 3 so 5!=120
  4 units place is divisable by either 2 or 6=(2*4!)=48
  5 units place is divisable by 5 so 5!=120
  6 units place is divisable by (3*2) so 5!=120
  so answer is 120+120+120+48+120+120=648
• 7 years agoHelpfull: Yes(5) No(1)
• option c ..
  if we take 1 at unti place then whole number is divisible by 1---so total combination ---5!=120
  if we take 2 at unti place then whole number is divisible by 2---so total combination ---5!=120
  if we take 5 at unti place then whole number is divisible by 5---so total combination ---5!=120
  if we take 6 at unti place then whole number is divisible by 6(bcz a num is divisible by 6 iff it iss divisible by 2 and 3..and if we take 6 at unit digit then it is divisible by 2 and rest of digit is 1+2+3+4+5=15 divisible by 3.(we can take it in any order))---so total combination ---5!=120
  if we take 3 at unti place then whole number is divisible by 3(if unit digit is either 3 or 6...and 6 we already considered so only option at unti digit is 3..)---so total combination ---5!=120
  if we take 4 at unti place then whole number is divisible by 4(then last 2 digit must be divisible by 4 ..and here at last 2 digit only 2 option 24 and 64)---so total combination ---4!*2=48
  .......................................................................
  total 648
• 7 years agoHelpfull: Yes(2) No(0)
• ans is 528
  1 present at unit place =120
  2 present at unit place=120
  3 presenr at unit place=120
  4 present at unit place=24+24=48
  5 present at unit place=120
  there are no such numbers which is divisible by 6
  ans =528
• 7 years agoHelpfull: Yes(1) No(8)
• coz 52 and 32 does not ends with 4
• 6 years agoHelpfull: Yes(1) No(0)
• First we fix 1 at unit place of a six digit number.
  _ _ _ _ _ 1
  _ _ _ _ _ 1
  _ _ _ _ _ 1
  _ _ _ _ _ 1
  _ _ _ _ _ 1
• 7 years agoHelpfull: Yes(0) No(7)
• when we put 1 at unit place- 5!
  when 2 -5!
  when 3 -5!
  when 4 -4!*2
  when 5 -5!
  when 6 -5!
  so total = 5*5!+4!*2=648
• 7 years agoHelpfull: Yes(0) No(1)
• guys all of you are taking 24 and 64 but 52 and 32 are also there what about them
• 6 years agoHelpfull: Yes(0) No(1)